Important Questions for CBSE Class 10 Science Chapter 11 - Electricity 2024-25

Very Short Answer Questions (1 Mark)

1. Which two circuit components are connected in parallel in the following circuit diagram? 

(a) \[{{\text{R}}_{\text{1}}}\] and ${{\text{R}}_{\text{2}}}$ only

(b) ${{\text{R}}_{\text{1}}}{\text{,}}{{\text{R}}_{\text{2}}}$ only 

(c) ${{\text{R}}_{\text{2}}}$ and ${\text{V}}$ only 

(d) ${{\text{R}}_{\text{1}}}$ and ${\text{V}}$only 

Ans: (a) The two circuit components that are connected parallel in the circuit diagram is  and ${R_2}$ only.

2. A metallic conductor has loosely bound electrons called free electrons. The metallic conductor is 

(a) negatively charged 

(b) positively charged 

(c) neutral 

(d) Either positively charged or negatively charged 

Ans: (c) The metallic conductor is neutral.

3. Which of the following expressions does not represent the electric power in the circuit? 

(a) ${\text{VI}}$.

(b) ${{\text{I}}^{\text{2}}}{\text{/R}}$

(c) ${{\text{V}}^{\text{2}}}{\text{/R}}$

(d)  ${{\text{I}}^{\text{2}}}{\text{R}}$

Ans: (b) The expression which does not represent the electric power in the circuit is ${I^2}/R$ .

4. Resistivity of a metallic wise depends on 

(a) its length 

(b) its shape 

(c) its thickness 

(d) nature of material 

Ans: (d) Resistivity of a metallic wire depends on the nature of the material.

5. If the current I through a resistor is increased by \[{\mathbf{100}}\% \] the increased in power dissipation will be (assume temperature remain unchanged) 

(a) \[{\mathbf{100}}\% \]

(b) \[{\mathbf{200}}\% \]

(c) \[{\mathbf{300}}\% \]

(d) \[{\mathbf{400}}\% \]

Ans: (c) The increase in power dissipation will be \[300\% \] .

6. For the circuit arrangement shown below, a student would observe.

(a) Some reading in both ammeter and voltmeter. 

(b) No reading in either the ammeter or the voltmeter. 

(c) Some reading in the ammeter but no reading in the voltmeter. 

(d) Some reading in the voltmeter but no reading in the ammeter. 

Ans: (c) A student will observe some reading in the ammeter but no reading in the voltmeter.

7. A wire of resistance $R$ is cut into five equal pieces. These pieces are connected in parallel and the equivalent resistances of the combination are $R'$ . Then the ratio $\dfrac{R}{{R'}}$ is 

(a) $\dfrac{{\text{1}}}{{\text{5}}}$

(b) ${\text{5}}$

(c) $\dfrac{{\text{1}}}{{{\text{25}}}}$

(d) ${\text{25}}$

Ans: (d)The ratio $\dfrac{R}{{R'}}$ is $25$ .

8. The resistance of the conductor is $R$ . If its length is doubled, then its new resistance will be 

(a) $R$

(b) $2R$

(c) $4R$

(d) $8R$

Ans: (c) The new resistance is $4R$ .

9. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances ${R_1},{R_2},{R_3}$ respectively as shown in the figure. Which of the following is live? 

(a) ${R_3} > {R_2} > {R_1}$

(b) ${R_2} > {R_3} > {R_1}$

(c) ${R_1} > {R_2} > {R_3}$

(d) ${R_1} = {R_2} = {R_3}$

Ans: (a)According to the graph , ${R_3} > {R_2} > {R_1}$

10. The nature of the graph between potential difference and the electric current flowing through a conductor is 

(a)parabolic 

(b) circle 

(c) straight line 

(d) hyperbolic 

Ans: (c)The nature of the graph between potential difference and the electric current flowing through a conductor is a straight line.

11. An electric heater is salted at $1500$ w. How much heat is produced per hour? 

(i) $5400$ J 

(ii) $54000$ J 

(iii) $5.4 \times {10^5}$

(iv) $5.4 \times {10^6}$.

Ans: (iv) The electric heater produces $5.4 \times {10^6}$ J per hour.

12. A student says that the resistance of two wires of the same length and same area of  cross section is the same. This statement is correct if 

(a) Both wires are of different materials 

(b) Both wires are made of the same material and are at different temperatures.

(c) Both wires are made of the same material and are at the same temperature.

(d) Both wires are made of different materials and are at the same temperature. 

Ans: This statement is correct if (c)The resistance of two wires of the same length and same area of cross section is the same if both wires are made of the same material and are at the same temperature.

13. In an experiment ohm, s law a student obtained a graph as shown in the diagram. The value of resistance of the resistor is 

(a) $0.1\Omega $

(b) $1.0\Omega $

(c) $10\Omega $

(d) $100\Omega $

Ans: (d) The value of resistance of the resistor is $100\Omega $ .

14. Work done to move $1$ coulomb charge from one point to another point on a charged conductor having potential $10$ volt is 

(a) $1$ Joule 

(b) $10$ Joule 

(c)  zero 

(d) $100$ Joule 

Ans: (c) Work done to move $1$ coulomb charge from one point to another point on a charged conductor having potential $10$ volt is zero. 

15. Three resistors are shown in the figure. The resistance of the combination is 

(a)$3\Omega $

(b) $6\Omega $

(c) $9\Omega $

(d) $7\Omega $

Ans: (c) The resistance of the combination is $9\Omega $ .

16. Name a device that helps to maintain a potential difference between across a conductor. 

Ans: A device that helps to maintain a potential difference between conductors is the battery.

17. What determines the rate at which energy is delivered by a current? 

Ans: The rate at which energy is delivered by a current is determined by electric power. 

18. A wire of resistance $R$ is cut into five equal pieces. These pieces are connected inparallel and the equivalent resistances of the combination are $R'$ . Then the ratio $\dfrac{R}{{R'}}$ is 

(a) $\dfrac{1}{5}$

(b) $5$

(c) $\dfrac{1}{{25}}$

(d) $25$

Ans: (d)A wire of resistance ${\text{R}}$ is cut into five equal pieces. These pieces are connected inparallel and the equivalent resistances of the combination are ${\text{R'}}$. In this cases, the ratio$\dfrac{R}{{R'}}$ is $25$ .

19. Which of the following terms does not represent electrical power in a circuit?

(a) ${I^2}R$

(b) $I{R^2}$

(c)  $VI$

(d) ${V^2}/R$

Ans: (b) The term that does not represent electrical power in a circuit is $I{R^2}$ .

20. An electric bulb is rated $220$ V and $100$ W. When it is operated on $110$ V, the power consumed will be: 

(a) $220$ W 

(b) $75$ W 

(c) $50$ W 

(d) $25$ W 

Ans: (d)The power consumed will be $25$ W.

21. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combination would be: 

(a) $1:2$

(b) $2:1$

(c) $1:4$

(d) $4:1$

Ans: (c) The ratio of heat produced in series and parallel combination will be $1:4$ .

22. A wire of resistance $R$ is bent in the form of a closed circle, what is the resistance across a diameter of the circle? 

Ans: The resistance across a diameter of the circle 

$ = R/4$

23. A charge of $6$ C is moved between two points $P$ and $Q$ having, potential $10$ V and $5$ V respectively. Find the amount of work done. 

Ans: The amount of work done, $W = q({V_2} - {V_1})\;$

$\; = {\text{ }}6\left( {10 - 5} \right)$

\[ = {\text{ }}30\] joule

24. Name the physical quantity whose SI unit is JC 

Ans: The physical quantity whose SI unit is JC is Potential.

25. Why are copper wires used as connecting wires? 

Ans: Copper wires are used as connecting wires because in case of copper the electrical resistivity for it is low.

26. A wire of resistivity $p$ is stretched to double its length. What is its new resistivity? 

Ans: When a wire of resistivity ${\text{p}}$ is stretched to double its length, then the new resistivity remains the same because resistivity depends on the nature of material.

27. What is the resistance of the connecting wire? 

Ans: The resistance of a connecting wire made of a good conductor is extremely low.

28. What is the resistance of an ammeter? 

Ans: An ammeter's resistance is very minimal, and in an ideal ammeter, it is zero.

29. What is the resistance of a Voltmeter? 

Ans: An ideal voltmeter's internal resistance is infinite.

Short Answer Questions (2 Marks)

1. How does use of fuse wire protect electrical appliances? 

Ans: When a large quantity of current passes through the circuit, the temperature of the wire rises and the fuse wire melts. This prevents current from flowing into the house's other circuits, saving electrical appliances.

2. Calculate the resistance of an electric bulb which allows a $10$A current when connected to a $220$ V power source? 

Ans: It is given from the question that an electric bulb allows a $10$A current when connected to a $220$ V power source.

Therefore,

\[I = 10\] A ,

\[V = 220\]V

\[R = \dfrac{V}{I}\]

\[ = \dfrac{{220}}{{10}}\]

\[ = 22\] ohm   

3. (i) Identify the $V - I$ graphs for ohmic and non-ohmic materials.

Ans: The $V - I$ graphs for ohmic and non-ohmic materials respectively can be represented as shown below:

(ii) Give one example of each.

Ans: (ii) Some examples of ohmic material are Copper, Nichrome and some examples of Non-ohmic material   are Diode, Transistor.

4. What do the following symbols represent in a circuit? Write the name and one function of each? 

(i)

Ans: (i) It symbolises a battery that maintains a potential difference across the circuit element to allow current to flow.

(ii)

Ans: It's an ammeter that measures how much current is flowing across a circuit.

5. Define the term “volt”? 

Ans: If $1$joule of energy is transferred between two points A and $B$, the potential difference between them is one volt. In an electric circuit, work is done to move one coulomb of charge from one point to another field.

6. Why does the connecting rod of an electric heater not glow while the heating element does? 

Ans: As its resistance is lower than that of the heating element, the connecting cord of an electric heater does not glow. As a result, the heating element produces more heat than the connecting cord, and it glows

7. A number of \[n\] resistors each of resistance \[R\] are first connected in series and then in parallel. What is the ratio of the total effective resistance of the circuit is series combination and parallel combination? 

Ans: Total effective resistance of  the circuit when in series combination \[{R_s} = nR\]

And for parallel combination is \[{R_p} = \dfrac{R}{n}\]  and 

$\dfrac{{{R_s}}}{{{R_p}}} = \dfrac{{nR}}{{\dfrac{R}{n}}}$

$ = {n^2}$

The ratio will be ${n^2}$ .

8. Draw a schematic diagram of a circuit consisting of \[3\] V battery, \[5\] ohm, \[3\Omega \]  and \[1\Omega \] resistor, an ammeter and a plug key, all connected in series. 

Ans: The circuit diagram of a circuit consisting of \[3\] V battery, \[5\] ohm, \[3\Omega \]  and \[1\Omega \] resistor, an ammeter and a plug key, all connected in series can be represented as show below,

9. A copper wire has diameter \[0.5\] mm and Resistivity of \[1.6 \times {10^{ - 8}}\Omega m\] What is the length of this wire to make its resistance ? How much does the resistance change if diameter is doubled?

Ans: Diameter of the copper wire, $D = 0.5 \times {10^{ - 3}}$ m

$P = 1.6 \times {10^{ - 8}}$

$R = 10$

Then,

$R = \dfrac{{\rho l}}{A}$

$ = \dfrac{{\rho l}}{{\pi {r^2}}}$

$ = \dfrac{{4\rho l}}{{\pi {D^2}}}$

$ \Rightarrow l = \dfrac{{\pi R{D^2}}}{{4\rho }}$

$l = \dfrac{{3.14 \times 10 \times {{(5 \times {{10}^{ - 4}})}^2}}}{{4 \times 1.62 \times {{10}^{ - 8}}}}$

$ = 121.14$

Length of the wire,  $l = 121.14$ m

New $R' = \dfrac{{4\rho l}}{{\pi {{(D')}^2}}}$

$ = \dfrac{1}{4}\dfrac{{4\rho l}}{{\pi {D^2}}}$

$ = \dfrac{1}{4}R$

Length of the wire to make its resistance $10\Omega $ is $121.14$m and when the diameter is doubled the new resistance will be one fourth that of the old one.

10. Alloys are used in electrical heating devices rather than pure metals. Give a reason. 

Ans: Alloys are utilised in electricity heating devices rather than pure metals because alloys have a higher resistivity and hence produce more heat. Furthermore, alloy is non-combustible (or oxidises easily at higher temperature).

11. On what factor does the resistance of a conductor depend? 

Ans: The factors on which Resistance depends are:-

(a) Length of the conductor

(b) Area of cross - section

(c)  Temperature

(d) Nature of material 

12. Calculate the number of electrons consisting of one coulomb of charge? 

Ans: Let $x = $ no. of electrons 

Charge on $1$ electron $ = 1.6 \times {10^{ - 19}}$ C , that is 

$x = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}$

$x = 6.25 \times {10^{18}}$

The number of electrons consisting of one coulomb of charge is $6.25 \times {10^{18}}$ .

13. What does an electric circuit mean? 

Ans: An electric circuit is a current route that is both continuous and closed. Current can flow through an electric circuit if it is complete.

14. Define the unit of current. 

Ans: The ampere is the SI unit for electric current. If $1$ coulomb charge flows per second across a conductor cross-section, the current is said to be $1$ ampere.

15. Calculate the number of electrons constituting one coulomb of charge. 

Ans: The charge on one electron $ = 1.6 \times {10^{ - 19}}$ coulomb.

Number of electrons in one coulomb of charge$ = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}$

$ = 6.25 \times {10^{18}}$

The number of electrons consisting of one coulomb of charge is $6.25 \times {10^{18}}$ .

16. What is meant by saying that the potential difference between two points is \[1\] v? 

Ans: If $1$ joule of labour is required to move a charge of $1$ coulomb from one location to another, the potential difference between the two points is said to be $1$ volt.

17. Ammeter burns out when connected in parallel. Give reasons. 

Ans: When a low-resistance wire is connected in series, a huge quantity of current travels through it, causing it to be burned, or short-circuited.

18. Judge the equivalent resistance when the following are connected in parallel: 

(a) Equivalent resistance of \[1\Omega \] and \[{10^6}\Omega \]

Ans: When the resistances are connected in a parallel arrangement, the resultant resistance is given by:

$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} +  \cdots $

$\dfrac{1}{R} = \dfrac{1}{1} + \dfrac{1}{{{{10}^6}}}$

$ = 1 + {10^{ - 6}}$

$R = 1\Omega $

The equivalent resistance is $1\Omega $ .

(b) Equivalent resistance of  \[1\Omega \] , \[{10^3}\Omega \] and \[{10^6}\Omega \]

Ans: $\dfrac{1}{R} = \dfrac{1}{1} + \dfrac{1}{{{{10}^3}}} + \dfrac{1}{{{{10}^6}}}$

$ = 1 + {10^{ - 3}} + {10^{ - 6}}$

$R = 1\Omega $

The equivalent resistance is $1\Omega $ .

19. An electric iron of resistance \[20\] takes a current of \[5\] A. Calculate the heat developed in \[30\] s.

Ans: Resistance of electric iron, $R = 20\Omega $ , current, $I = 5$ A and time $ = 30$ s.

Heat generated $H = {I^2}Rt$

$ = {5^2} \times 20 \times 30$

$ = 15000$ j

The heat developed in $30$ second is $15000$ j .

20. Compute the heat generated while transferring \[96000\] coulomb of charge in one hour through a potential difference of \[50\] V

Ans: The given information is as shown below,

Charge transferred, $Q = 96000$

Potential Difference, $V = 50$  V.

Heat generated, $H = VQ$

$ = 50 \times 96000$

$ = 4800000$ j 

$ = 4.8 \times {10^6}$ j

The heat generated while transferring $96000$  coulomb of charge in one hour

through a potential difference of $50$ V is $4.8 \times {10^6}$ j

21. An electric motor takes \[5\] A from a \[220\] V line. Determine the power of the motor and energy consumed in \[2\] h. 

Ans: Given that current drawn by electric motor $I = 5$ A.

The line voltage $V = 220$ V 

Time, $t = 2$ h

 Power of motor , $P = VI$

$ = 220 \times 5$

$ = 1100$ W and 

the energy consumed $E = Pt$

$ = 2 \times 1100$

$ = 2.2$ KWh

The power of the motor and energy consumed in $2$ h are $1100$ W and $2.2$ kWh respectively.

22. How is a voltmeter connected in the circuit to measure the potential difference between two points? 

Ans: A voltmeter is connected in parallel to the resistance across the place where the potential difference is to be determined.

23. When a \[12\] v battery is connected across an unknown resistor, there is a current of \[2.5\] mA in the circuit. Find the value of the resistance of the resistor. 

Ans: Given that Voltage of battery, $V = 12$  V, 

Current, $I = 2.5$ mA 

$ = 2.5 \times {10^{ - 3}}$ A

Resistance, $R = V/I$

$ = \dfrac{{12}}{{2.5 \times {{10}^{ - 3}}}}$

$ = 4800\Omega $

The value of the resistance of the resistor is $4800\Omega $ .

24. Several electric bulbs designed to be used on a \[220\] V electric supply line, are rated \[10\] W. How many lamps can be connected in parallel with each other across the two wires of \[220\] V line if the maximum allowable current is \[5\] A? 

Ans: The given information is as shown below

Each bulb is rated as $10$ W, $220$ V, 

It draws current, $I = P/V$

$ = \dfrac{{10}}{{220}}$ V

$ = 1/22$ A.

The maximum allowable current is $5$ A and all lamps are connected in parallel. Therefore the maximum number of bulbs joined in parallel with each other $ = 5 \times 22$whichis$110$.

25. Two lamps, one rated \[100\] W at \[220\] V, and the other \[60\] W at \[220\] V are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is \[220\] V? 

Ans: Current drawn by ${1^{st}}$lamp rated $100$ W at $220$ ,  $V = P/V$

\[ = 100/220\]

\[ = 5/11\] A. 

Current drawn by ${2^{nd}}$lamp rated $60$ W at \[220\] , \[V = 60/220\]\[ = 3/11\;\]A.

In parallel arrangement the total current \[ = {\text{ }}3/11 + {\text{ }}5/11\]

\[ = {\text{ }}8/11\]

\[ = {\text{ }}0.73\] A.

Current drawn from the line if the supply voltage is $220$ V is $0.73$ A .

26. Which uses more energy, a \[250\] W TV set in \[1\] hour, or a \[1200\] W toaster in \[10\] minutes? 

Ans: Energy used by a TV set of power $250$ W in $1$ hour $ = Pt$

$ = 250$Wh

Energy used by toaster of power \[1200\] W in \[10\] minute \[\left( {10/60h} \right) = {\text{ }}200\] Wh.

A $250$ W TV set in $1$ hour uses more energy.

27. An electric heater of resistance \[8\] draws \[15\] A from the service mains for \[2\] hours. Calculate the rate at which heat is developed in the heater. 

Ans: Resistance of electric heater, \[R{\text{ }} = {\text{ }}8\Omega \], 

current, \[I{\text{ }} = {\text{ }}15\] A.

Rate at which heat developed in the heater $ = \dfrac{{{I^2}Rt}}{t}$

$ = 1800$ W.

The rate at which heat is developed in the heater is $1800$ W.

28. In the given figure what is the ratio of current in  \[A\]

Ans: Observe that it I clearly known to us that $V = IR$

$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{R}{{2R}}$

$ = \dfrac{1}{2}$

The ratio of current is$1/2$.

29. Two wires of equal cross sectional area, one of copper and other of managing have the same resistance. Which one will be longer? 

Ans: Using the equation  $\rho  = \dfrac{{RA}}{l}$ , where $\rho $ is the resistivity, $R$ is the resistance and $A$ the area

We can see that copper wire has a lower resistance than manganin, hence copper will last longer.

30. A Rectangular block of iron has dimensions $L \times L \times b$ . What is the resistance of the block measured between the two square ends? Given $p$ resistivity. 

Ans: $R = \dfrac{{pb}}{{{L^2}}}$is the resistance of the block measured between the two square ends

31. Three equal resistances are connected in series then in parallel. What will be the ratio of their Resistances? 

Ans: ${R_{series}} = 3R$

${R_{parallel}} = R/3$

The ratio of Resistances is $9$ .

32. Justify for any pair of resistance the equivalent resistance equivalent resistance in parallel.

Ans: As $R = V/I$

From the graph for any pair of resistance the equivalent resistance in series is greater than equivalent resistance in parallel.

$A = $ Series, $B = $ Parallel 

33. How many bulbs of \[81\] should be joined in parallel to draw a current of \[2\] A from a battery of \[4\] V? 

Ans: $R = V/I$

$ = 4/2$

$ = 2\Omega $

let $n$ be the number of bulbs.

$1/R = 1/{R_1} + 1/{R_2} +  \cdots  + 1/{R_n} = \dfrac{n}{8}$

$ \Rightarrow \dfrac{1}{2} = \dfrac{n}{8}$

$ \Rightarrow n = 4$

Number of bulbs are $4$ .

34. Two cubes \[A\] and \[B\] are of the same material. The side of \[B\] is thrice as that of \[A\] . Find the ratio \[{R_A}/{R_B}\] . 

Ans: The value of  ${R_A} = \dfrac{{pL}}{A}$ and

${R_B} = \dfrac{{p3L}}{{9A}}$

${R_A}:{R_B} = 3:1$

35. If there are $3 \times {10^{11}}$ electrons are flowing through the filament of the bulb for two minutes. Find the current flowing through the circuit. Charge on one electron $1.6 \times {10^{19}}$

Ans: Observe as shwn below,

Using the equation

$q = ne$

$ = 3 \times {10^{11}} \times 1.6 \times {10^{19}}$C

$ = 4.8 \times {10^8}$C

$I = q/t$

$ = \dfrac{{4.8 \times {{10}^8}}}{{2 \times 60}}$

$ = 4 \times {10^7}$ A

The current flowing through the circuit is $4 \times {10^7}$ A.

36. A nichrome wire of resistivity $100$ W m and copper wire of resistivity $1.62$ M ohm-m of the same length and same area of cross section are connected in series , current is passed through them, why does the nichrome wire get heated first? 

Ans:  Looking at the equation

$Q = {I^2}RT$

$Q = {I^2}(pL/A)t$

Henceforth, because nichrome wire has a higher resistance than copper wire, it must be heated first.

37. What is represented by joule/coulomb? 

Ans: The potential difference is represented by the joule/coulomb.

38. A charge of $2$ C moves between two plates, maintained at a p.d of $IV$ . What is the energy acquired by the charge? 

Ans: The energy acquired by the charge, $W = QV$

$ = 2$J

The energy acquired is $2$ J

39. Which has more resistance: $100$ W bulb or $60$ W bulb? 

Ans: As, it is clearly known that$R \propto \dfrac{1}{P}$, thus the resistance of $60$ W bulb is more.

40. What happens to the current in a circuit if its resistance is doubled? 

Ans: As current and resistance are inversely proportional, the current is reduced to half of its previous value.

41. What happens to the resistance of a circuit if the current through it is doubled? 

Ans: Resistance is unaffected since the circuit's resistance is independent of the current flowing through it.

42. How does the resistance of a wire depend upon its radius? 

Ans: As $R \propto \dfrac{1}{A}$

$ \Rightarrow R \propto V$

Resistance of a wire is directly proportional to its radius.

43. Two wires are of the same length, same radius, but one of them is of copper and the other is of iron. Which will have more resistance. 

Ans: Since  \[R{\text{ }} = {\text{ }}p1/A\] ,

but A and I are the same. It is solely determined by resistivity, hence iron has a higher resistance.

44. Two wires of same material and same length have radii $R$ and $r$ Compare their resistances. 

Ans: Suppose $R$ and $r$ are resistances, then $R = r$ as $p$ and $I$ are the same.

Short Answer Questions (3 Marks)

1. Two metallic wires A and B are connected, wire A has length I and radius $r$ , while  wire B has  length  $2l$  and  radius $2r$ .  Find  the  ratio  of  total  resistance  of  series combination and the resistance of wire $A$ , if both the wires are of the same material? 

Ans: Observe as shown below,

Resistance of metallic wire $A$ , ${R_1} = \dfrac{{\rho l}}{A}$

$ = \dfrac{{\rho l}}{{\pi {r^2}}}$

Resistance of metallic wire $B$ , ${R_2} = \dfrac{{\rho 2l}}{{4\pi {r^2}}}$

Total resistance in series is $R = {R_1} + {R_2}$

$ = \dfrac{{\rho l}}{{\pi {r^2}}} + \dfrac{{2\rho l}}{{4\pi {r^2}}}$

$ = \dfrac{{3\rho l}}{{2\pi {r^2}}}$

The ratio of the total resistance is series to the resistance of $A$ is

$\dfrac{R}{{{R_1}}} = \dfrac{{\dfrac{{\rho l}}{{\pi {r^2}}}}}{{\dfrac{{3\rho l}}{{2\pi {r^2}}}}}$

$ = 2/3$

The ratio of the total resistance is series to the resistance of $A$ is $2/3$ .

2. Should  the heating  element  of  an  electric  iron be made  of  iron,  silver  or nichrome wire? Justify giving three reasons? 

Ans: The following reasons can be found, why the heating element of an electric iron is composed of nichrome wire.

(1) Due to the high resistance, the passage of current generates additional heat.

(2) High melting point.

(3)  At high temperatures, it does not easily oxidised (or burn).

3. (a) Define electric resistance of a conductor? 

Ans: A conductor's electric resistance is defined as the resistance it provides to the flow of current.

That is $R = V/I$ and its S.I. unit is ohm , $\Omega $ .

(b) A wire of  length $L$ and resistance $R$  is stretched so that  its  length  is doubled and the area of the cross section is halved. How will its 

(i) resistance change 

Ans: It is clearly known that resistance,  $R = \dfrac{{\rho l}}{A}$

New length $L' = 2L$ and

$A' = \dfrac{A}{2}$

Therefore ${R^1} = \dfrac{{\rho L'}}{{A'}}$

$ = 4R$

Therefore, the resistance of a wire becomes 4 times its original resistance.

(ii) resistivity change? 

Ans: The size of a wire has no bearing on its resistance. As a result, resistance does not vary.

4. Two  resistors  of  resistance  $R$  and  $2R$  are  connected  in parallel  in  an  electric  circuit. Calculate the ratio of the electric power consumed by $R$ and $2R$ ? 

Ans: Power consumed by $R$ , ${\rho _1} = \dfrac{{{V^2}}}{R}$

Power consumed by$2R$ , ${\rho _2} = \dfrac{{{V^2}}}{{2R}}$

Ratio $\dfrac{{{\rho _1}}}{{{\rho _2}}} = \dfrac{{\dfrac{{{V^2}}}{R}}}{{\dfrac{{{V^2}}}{{2R}}}}$

.$ = 2:1$.

The ratio of the electric power consumed by $R$ and $2R$ is $2:1$ .

5. The length of different metallic wires but of same area of cross section and made of the same material are given below 

(i) Out of these two wires which wire has higher resistance. 

Ans: As $R \propto l$ (length of the conductor) and since length of wire $C$ is more than $A$ and $B$ ,  wire $C$ has higher resistance.

(ii) Which wire has higher electrical resistance? Justify your answer. 

Ans: The electrical resistivity of a wire is determined by the nature of the material, not by its dimensions. As a result, the resistivity of all wires is the same as the substance of all wires.

6. Two resistors of resistances $R$ and $2R$ are connected in series with an electrical circuit? Calculate the ratio of the electric power consumed by $R$ and $2R$ ? 

Ans: It is clearly known thatElectric power consumed by $R$ , ${P_1} = {I^2}R$

Also, electric power consumed by $2R$ , ${P_2} = {I^2}2R$

$\dfrac{{{P_1}}}{{{P_2}}} = 1/2$

The ratio of the electric power consumed by $R$ and $2R$ is$1:2$.

7. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. the ratio of heat produced in series and parallel combinations would be 

(a) $1:2$

(b) $2:1$

(c) $1:4$

(d) $4:1$

Ans: (c)Let resistance of each wire is $R$

In series, resistance is  $ = 2R$

Heat produced, ${H_1} = \dfrac{{{V^2}}}{{2R}}t$

In parallel total resistance $ = R/2$

Heat produced, ${H_2} = \left( {\dfrac{{{V^2}}}{{\dfrac{R}{2}}}} \right)t$

$ = \dfrac{{2{V^2}}}{R}t$

${H_2} = 4{H_1}$

$\dfrac{{{H_1}}}{{{H_2}}} = 1/4$

The ratio of heat produced in series and parallel combinations is $1:4$ .

8. Calculate the following of a circuit shown in the figure.

(i) effective resistance 

Ans: Effective resistance, $R = {R_1} + {R_2}$

$ = 5 + 10$

$ = 15$

The effective resistance is $15\Omega $ .

(ii) current 

Ans: Current, $I = V/R$

$ = 2/15$

$ = 0.133$

The current is $0.133$ A.

(iii) Potential difference across $10\Omega $ resistor

Ans: Potential difference across $10\Omega $

$V = IR$

$ = \dfrac{2}{{15}} \times 10$

$ = 1.33$

The potential difference across $10\Omega $ is $1.33$ volt.

9. A Piece of wire of resistance $20\Omega $ is drawn out so that its length is increased to twice its original length to calculate the resistance of the wire is the new situation? 

Ans: Resistance of wire$ = 20\Omega $

As $R = \dfrac{{\rho l}}{A}$

And as length of a wire is increased, its area of cross- section decreases, and volume of the wire remains constant.

$l' = 2l$

$A' = A/2$

$R' = \dfrac{{\rho l'}}{{A'}}$

$ = \dfrac{{4\rho l}}{{A'}}$

$\dfrac{{R'}}{R} = \dfrac{4}{1}$

$ \Rightarrow R' = 80$

The resistance of the wire is the new situation is $80\Omega $ . 

10. A  battery  made  of  $5$  cells  each  of  $2$ V  and  have  internal  resistance $0.1\Omega ,0.2\Omega ,$$0.3\Omega ,0.4\Omega $ and \[0.5\Omega \]  is  connected  across \[10\Omega \] resistance.  Draw a circuit diagram and calculate the current flowing through$10\Omega $resistance?

Ans: The internal resistance , \[0.1 + 0.2 + 0.3 + 0.4 + 0.5 = 1.5\Omega \]

Total resistance \[ = 1.5 + 10\]

\[ = 11.5\]

\[I = V/R\]

\[ = 10/11.5\]

\[I = {\text{ }}0.869\] A

Current flowing through $10\Omega $ resistance is $0.869$ A.

11. In the circuit diagram given here Calculate- 

(a) The total effective resistance 

Ans: As resistances are in parallel

$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}$

$ = \dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{{10}}$

$ = \dfrac{8}{{10}}$

$R = \dfrac{{10}}{8}$

Total effective resistance is $\dfrac{{10}}{8}\Omega $

(b) The total current 

Ans: Total current, $I = V/R$

$ = \dfrac{6}{{10/8}}$

$ = 4.8$

The total current is $4.8$ A.

(c) The current through each resistor. 

Ans: From the circuit diagram, ${I_1},{I_2}$ and ${I_3}$ be the current through $2\Omega ,5\Omega $ and $10\Omega $  respectively

Therefore, ${I_1} = \dfrac{V}{{{R_1}}}$

$ = 3$

${I_2} = \dfrac{V}{{{R_2}}}$

$ = 1.2$

${I_3} = \dfrac{V}{{{R_3}}}$

$ = 0.6$

Current through each$2\Omega ,5\Omega $ and $10\Omega $  is $3$ A ,$1.2$ A, and $0.6$ A respectively.

12. You have two circuits Compare the power used in \[2\Omega \] resistor in each case. 

(i) a \[6V\] battery is series with \[1\Omega \] and \[2\Omega \] resistor

Ans: Potential difference, $V = 6V$

${R_1} = 1\Omega $

${R_2} = 2\Omega $

Total Resistor $ = {R_1} + {R_2}$

$ = 1 + 2$

$ = 3$

$I = \dfrac{V}{R}$

$ = \dfrac{6}{3}$

$ = 2$

${P_1} = \dfrac{{{V^2}}}{R}$

$ = 8$

The power used in $2\Omega $resistor  is $8$ W.

(ii) a \[4V\] battery in parallel with  \[12\Omega \] and \[2\Omega \] resistor 

Ans: Potential difference, $V = 4V$

${R_1} = 12\Omega $

${R_2} = 2\Omega $

${P_2} = \dfrac{{{V^2}}}{R}$

$ = 8$

The power used in $2\Omega $resistor  is $8$ W.

The ratio of both power is $1:1$ .

13. How  much energy  is  given  to  each coulomb  of charge  passing  through  a  \[6\]  volt battery? 

Ans: Potential difference, $V = 6V$

 Charge, $Q = 1C$

Energy = total work done  

\[ = {\text{ }}Q{\text{ }}x{\text{ }}V\]

\[ = {\text{ }}1x6\]

\[ = {\text{ }}6\] joule.

Energy  given to each coulomb of charge passing through a $6$ volt battery is $6$ joules.

14. On what factor does the resistance of a conductor depend? 

Ans: A conductor's resistance is determined by the following factors:

(i) length of conductor

(ii) Area of cross-section

(iii) Temperature

(iv) Conductors are made from a variety of materials.

15. Will  current  flow  more  easily  through  a  thick  wire  or  a  thin  wire  of  the  same material, when connected to the same source? Why? 

Ans: When linked to the same source, current flows more freely through a thick wire than via a small wire of the same material. It's because resistance rises as thickness decreases.

16. Let  the  resistance  of  an  electric  component  remain  constant  while  the  potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it? 

Ans: The electrical component's resistance R remains unchanged, but the potential difference across its ends falls to half of its original value. As a result of Ohm's law, new current is reduced to half of its initial value.

17. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal? 

Ans: The following reasons are why the coils of electric toasters and electric irons are built of an alloy rather than a pure metal:

(i) An alloy's resistivity is higher than that of pure metal.

(ii) An alloy does not rust quickly at high temperatures.

18. Draw a schematic diagram of a  circuit  consisting of a battery of  three  cells of \[2V\] , each, a \[5\Omega \] resistor, \[8\Omega \] resistors and a \[12\Omega \] and a plug key, all connected in series. 

Ans: The diagram of circuit is as follows

19. An electric lamp of \[100\Omega \] , a toaster of resistance \[50\Omega \] and a water filter of resistance \[500\Omega \] are connected in parallel to a \[220V\] source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it? 

Ans: It can be found from the question, voltage, $V = 220V$

${R_1} = 100\Omega $

${R_2} = 50\Omega $

${R_3} = 500\Omega $

\[1/R = 1/100 + 1/50 + 1/500\]

\[R{\text{ }} = {\text{ }}500/16\]

\[ = {\text{ }}31.25\Omega \]

The resistance of electric iron, which draws as much current as all three appliances take together is $31.25\Omega $ . 

Current passing through electric iron,  \[I = V/R\]

\[ = 220/31.25\]

\[ = 7.04\] A.

That is the current passing is $7.04$ A.

20. What  is (a) the  highest total  resistance  that  can  be  secured  by a combination of four resistance of \[4\Omega ,8\Omega ,12\Omega \] and \[24\Omega \] ?

Ans: When all four resistances must be connected in series, the highest resistance is achieved. In that instance, the outcome

$R = {R_1} + {R_2} + {R_3} + {R_4}$

\[ = {\text{ }}4 + 8 + 12 + 24\]

\[ = {\text{ }}48\Omega \]

The highest resistance is \[48\Omega \] .

(b) The  lowest  total  resistance  that  can  be  secured  by a combination of four resistance of \[4\Omega ,8\Omega ,12\Omega \] and \[24\Omega \] ?

Ans: All four resistances must be connected in parallel to produce the lowest resistance.

$1/R = 1/{R_1} + 1/{R_2} + 1/{R_3} + 1/{R_4}$

\[ = {\text{ }}1/4{\text{ }} + 1/8{\text{ }} + 1/12{\text{ }} + {\text{ }}1/24\]

\[ = {\text{ }}12/24\]

The lowest resistance is $2\Omega $ .

21. Why does the cord of an electric heater not glow while the heating element does? 

Ans: When connected to the voltage source, the cord of a heater and the cord of an electric heater are connected in series and carry the same current.

Because the resistance of the cord is so low in comparison to the resistance of the heater element.

As a result, the amount of heat created in the cord is extremely low, but significantly higher in the heater element. As a result, the heating element glows, but the cord does not.

22. A copper wire has diameter \[0.5\] mm and resistivity of \[1.6 \times {10^{ - 8}}\] m. What will be the length of  this wire  to make  its resistance \[10\] ? How much does  the resistance change  if the diameter is doubled? 

Ans: The diameter of wire, d = 0.5 mm, 

Resistivity, $\rho  = 1.6 \times {10^{ - 8}}$

resistance  $R = {\text{ }}10{\text{ }}\Omega $

\[R = \rho L/A\]

$L = \dfrac{{\pi {D^2}R}}{{4\rho }}$

\[ = \dfrac{{22 \times {{(5 \times {{10}^{ - 4}})}^2}}}{{7 \times 4 \times 1.6 \times {{10}^{ - 8}}}}\]

\[ = {\text{ }}122.5\] m

As resistance is inversely proportional to the cross-section area of wire, when the diameter is doubled for a given length of material, the resistance reduces.

23. A battery of \[9V\] is connected in series with resistance of \[0.2\Omega ,0.3\Omega ,0.4\Omega ,0.5\Omega \] and \[12\Omega \]respectively. How much current would flow through the \[12\] resistor? 

Ans: Potential difference $V = 9V$ . 

Total resistance   \[ = {\text{ }}0.2{\text{ }} + 0.3{\text{ }} + {\text{ }}0.5{\text{ }} + {\text{ }}0.5{\text{ }} + {\text{ }}12{\text{ }}\]

\[ = {\text{ }}13.4{\text{ }}\Omega \]

Current in the circuit \[I{\text{ }} = {\text{ }}V/R{\text{ }}\]

\[ = {\text{ }}9{\text{ }}V{\text{ }}/{\text{ }}13.{\text{ }}4{\text{  }}\] A. 

\[ = {\text{ }}0.67\]

In a series circuit the same current  flows  through all  the resistance, hence current of \[0.67\] A will flow through \[12{\text{ }}\Omega \] resistor. 

24. How many \[176\Omega \] resistors (in parallel) are required to carry \[5\] A on a \[220\] V line? 

Ans: Let the resistors of \[176\Omega \] be joined in parallel. 

Their combined resistance,

\[1/R = 1/176 + 1/176 \ldots  \ldots \] times

\[ = n/176\;\] or

$ \Rightarrow R = 176/n\Omega $

Given that \[V = 220V\] and 

\[I = 5\] A 

\[R = V/I\]

\[ = 176/n\]

\[ = 220/5\]

\[ = 44\Omega \;\;\;\;\;\;\]

\[n{\text{ }} = {\text{ }}176/44{\text{ }}\]

\[ = {\text{ }}4\] , 

A number of $4$ resistors should be joined in parallel.

25. Show  how  you  would  connect  three  resistors,  each  of  resistance  \[6\Omega \]  so  that  the combination has resistance of 

(i) \[9\Omega \]

Ans: From the question, ${R_1} = {R_2}$

$ = 6$

Join three resistors as below to get net resistance of $9\Omega $ :

(ii) \[4\Omega \]

Ans: Join three resistors as below to get \[4\Omega \] net resistance :

26. A hot plate of an electric oven connected to a \[220V\] line has two resistance coils \[A\] and \[B\] . Each of \[24\Omega \] resistances, which may be used separately, in series or in parallel. What are the currents in the three cases? 

Ans: From the question, potential difference\[V = {\text{ }}220{\text{ }}V\] .

Resistance of coil $A = $ Resistance of coil \[B\]

\[ = 24{\text{ }}\Omega \]

When coil is used separately, the circuit \[I = V/R\]

\[ = 220V/24\Omega \]

\[ = {\text{ }}9.2{\text{ }}A\]

The current is\[9.2{\text{ }}A\].

When coils are used in series total resistance $R = {R_1} + {R_2}$

\[ = 24 + 24\]

\[ = 48\Omega \]

The current flowing, \[I = V/R\]

\[ = 220V/48\Omega \]

\[ = 4.6\] A

The current is $4.6$ A.

Two coils are joined in parallel. 

Total resistance \[R = {\text{ }}1/24{\text{ }} + {\text{ }}1/24\]

\[ = {\text{ }}2/24\]

\[R = 12{\text{ }}\Omega \] .

Current \[I = V/R\]

\[ = 220V/12\Omega \]

\[ = 18.3\] A.

The current is $18.3$ A.

27. Compare the power used in the \[2\Omega \] resistor in each of the following circuits: 

(i) a \[6\] volt battery in series with \[1\Omega \] and \[2\Omega \] resistors and, 

Ans: Suppose a \[2\Omega \] resistor is joined to a $6$ V battery in series with \[1\Omega \] and \[2\Omega \] resistors.

Total resistance \[R = 2 + 1 + 2\]

\[ = 5\Omega \]

Current \[I = 6V/5\Omega \]

\[ = 1.2\] A

Power used in $2A$ resistor \[ = {I^2}R\]

\[ = 2.88\] W

Power used is $2.88$ W.

(ii) a \[4V\] battery in parallel with \[12\Omega \] and \[2\Omega \] resistors. 

Ans: Suppose \[2\Omega \] resistor is joined to a $4V$ battery in parallel with \[12\Omega \] resistor and \[2\Omega \] resistors, 

the current flowing in \[2\Omega  = 4V/2\Omega \]

\[ = 2A\]

Power used in \[2\Omega \] resistor \[ = {I^2}R\]

\[ = 8\] W 

Ratio \[ = 2.88/8\]

\[ = 0.36:1\;\]

The ratio of power used is \[0.36:1\] .

28. In  the given  figure what  is  ratio of ammeter  reading when \[J\] is connected  to \[A\] and then to \[B\]

Ans: Connect  $J$  to $A$  ,     then

\[I = V/R\]

\[ = 3/5\]

\[ = 0.6A\]

When $J$ is connected to $B$\[V = 1 + 2 + 3 + 4\]

\[ = 10V\]

$I = 10/5$

\[ = 2A\]

Ratio of ammeter reading when J is connected to $A$ and then to $B$ is $3:10$ .

29. Given  a  resistor  each  of  resistors  \[R\] .  How  will  you  combine  them  to  get  the 

(i) maximum effective resistance?  

Ans: For maximum resistance $R = nr$  , this is the same as combining a series of numbers.

(ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance? 

Ans: For minimum resistance $R' = r/n$  , this is the same as combining a series of numbers.The ratio of the maximum to minimum resistance is $R/R' = {n^2}$.

30. A wire of  length \[L\] and resistance \[R\]  is stretched so that  its  length  is doubled. How will its

(a) Resistance change 

Ans: The resistance of a wire is determined by its length, cross-sectional area, and resistivity as\[{\text{R = }}\dfrac{{{{\rho l}}}}{{\text{A}}}\]

Hence, if the length is doubled and area is halved, then we have

$\dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{\dfrac{{\rho {l_2}}}{{{A_2}}}}}{{\dfrac{{\rho {l_1}}}{{{A_1}}}}}$

$ = \dfrac{{{l_2}{A_1}}}{{{l_1}{A_2}}}$

$ = 4$

Therefore, ${R_2} = 4{R_1}$

Hence, resistance of the wire becomes four times the original value.

(b) Resistively change?

Ans: The substance from which wire is formed has a property called wire resistivity. As a result, changing the wire's size has no effect on its resistivity.

31. Two students perform the experiments on series and parallel combinations of two given resistors \[{R_1}\]  and \[{R_2}\] and plot the following \[V - I\] graphs. 

Ans: Both students are  correct  because  

\[AV/A1 = \;\] resistance $R$  and  

\[A1/AV = l/R\;\]

The term "series" refers to high resistance, while "parallel" refers to low resistance.

32. A household uses the following electric appliances. Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. \[3.00\]

(i) Refrigerator of rating \[4\] for ten hours each day.

Ans: Month of June has \[30\] days. 

Refrigerator of $400$ W is running $2$ hours each day.

Total hours it is run in $30$ days \[ = 2 \times 30\]

\[ = 60\] h

Energy consumed in kWh is \[ = 400{\text{ }} \times {\text{ }}60/1000{\text{ }}\]

\[ = {\text{ }}24\] kWh

(ii) Two electric fans of rating \[80\]each for twelve hours each day. 

Ans: Two electric fans of \[80\] W are run \[12\] hours each day.

Total hours they are run in \[30\] days \[ = 12 \times 30\]

\[ = 360\]

Energy consumed in kWh is \[ = 2 \times 80 \times 360/1000\]

$ = 57.6$ kWh

(iii) Six electric tubes of rating \[18\] W each for 6 hours each day. 

Ans: Six electric tubes each of \[18\] W are run $6$ hours daily.

Total hours it is run in $30$ days \[ = 6 \times 30\]

\[ = 180\] h

Energy consumed in kWh is \[ = 6 \times 18 \times 180/1000\]

\[ = 19.44\]  kWh

Net energy consumed in the month of June is \[ = 24 + 57.6 + 19.44\]

\[ = 101.04\] kWh 

Thus, the electric bill is \[ = 3 \times 101.04\]

\[ = Rs303.12\]

Long Answer Questions (5 Marks)

1. Two wires  \[A\]  and  \[B\]  are  of  equal  length,  different  cross  sectional  areas  and made  of the same metal. 

(a) (i) Name the property which is same for both the wires, 

Ans: Resistivity - As resistivity is a property of a substance, it is constant for both wires.

(ii) Name the property which is different for both the wires. 

Ans: Resistances - As the cross sectional areas of each wires are different, they are treated as separate objects.

(b) If the resistance of wire \[A\] is four times the resistance of wire \[B\] , calculate 

(i) the ratio of the cross sectional areas of the wires and 

Ans: Since \[R = \dfrac{{\rho l}}{A}\]

For wire $A$ , ${R_1} = \dfrac{{\rho l}}{{{A_1}}}$

For wire $B$ , ${R_2} = \dfrac{{\rho l}}{{{A_2}}}$

\[ \Rightarrow \dfrac{{{R_2}}}{{{R_1}}} = \dfrac{{{A_1}}}{{{A_2}}}\]

Since ${R_1} = 4{R_2}$

$ \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = 1:4$

\[\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\pi {r_1}^2}}{{\pi {r_2}^2}}\]

\[ = {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}\]

Ratio is \[{\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2}\] .

(ii) The ratio of the radii of the wire. 

Ans: \[{\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^2} = 1/4\]

Ratio is $1:2$ .

2. (a) State ohm’s law? 

Ans: If the temperature and other physical conditions of the conductor stay constant, the electric current flowing through the conductor is precisely proportional to the potential difference across the conductor's end.

(b) The value of \[(I)\] current flowing through a conductor for the corresponding values of \[(V)\] potential difference are given below 

Plot a graph between \[V\] and \[I\] and also calculate resistance. 

Ans: Along $x$ -axis \[IV = 1\] cm 

$R = 5/3$

$ = 1.67$

The resistance  is $1.67\Omega $ .

3. (a) Define electrical energy with S.I. unit? 

Ans: The effort done by a source of electricity to sustain current in a circuit is known as electrical energy. The joule is its SI unit.

(b) A household uses the following electric appliance; Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. \[3.00\] . 

(i) Refrigerator of rating \[400\] w for ten hour each day. 

Ans: Electricity consumed by refrigerator in one day $ = $ power $ \times $ time

\[ = 400 \times \;10\]

\[ = {\text{ }}4000\] Wh

\[ = 4\] kwh

Therefore the electricity consumed is $4$  KWh. 

(ii) Two electric fans of rating \[80\] w each for twelve hours each day. 

Ans: Electricity consumed by $2$ electric fans in $1$ day, power $ \times $ time

\[ = 2 \times 80 \times 12\]

\[ = {\text{ }}1.92\] kwh

Therefore the electricity consumed is $1.92$ KWh. 

(iii) Six electric tubes of rating \[18\] w each for 6hours each day. 

Ans: Electricity consumed by $6$ electric tubes in $1$ day \[ = 6 \times 18 \times \;6\]

\[ = {\text{ }}0.648\] kwh

Therefore the electricity consumed is $0.648$ KWh. 

Total energy consumed in one day \[ = 4 + 1.92 + 0.648\]

\[ = 6.548\] kwh

Total energy consumed in one month \[ = 6.568 \times 30\]

\[ = 197.04\] kwh

Cost of $1$ unit (kwh) $ = $ Rs \[3.00\]

Cost of \[197.04\] kwh \[ = 197.04 \times 3\]

Electricity bill \[ = Rs591.12\]

The electricity bill of the household for the month of June is Rs. $591.12$.

4. Redraw  the  circuit  of  question \[1\] ,  putting  in  an  ammeter  to  measure  the  current through the resistors and a voltmeter to measure the potential difference across the \[12\Omega \] resistors. What would be the reading in the ammeter and voltmeter? 

Ans:

Ammeter $A$ has been joined in series with circuit and voltmeter $V$ is joined in parallel to $12$ ohms resistor.

Total voltage of battery \[V = 3x2\]

\[ = 6\] V.

Total resistance \[R = {R_1} + {R_2} + {R_3}\]

\[ = 5\Omega  + 8\Omega  + 12\Omega \]

\[ = 25\Omega \]

Ammeter reading \[ = I\]

\[ = V/R\]

\[ = 6/25\]

\[ = 0.24\] A.

Voltmeter reading \[ = {\text{ }}IR\]

\[ = 0.24{\text{ }}x{\text{ }}12\]

\[ = 2.88\] V.

The reading in the ammeter and voltmeter is $0.24$ A and $2.88$ V respectively.

5. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series? 

Ans: The following are the benefits of connecting electrical equipment in parallel with the battery rather than in series:

(i) Each connecting electrical device will have the same voltage, and the device will take current according to its resistance.

(ii) It is possible to use separate on/off switches.

(iii) As the total resistance in the parallel circuit falls, a large current can be pulled from the cell.

(iv) Even if one electrical gadget is broken, other devices continue to function normally.

6. How  can  three  resistors  of  resistance  \[2\Omega ,3\Omega \]and  \[6\Omega \]  be  connected  to  give  a  total resistance of

(a) \[4\Omega \]

Ans: If we connect resistance of $3\Omega $ and $6\Omega $ in parallel and resistance of $2\Omega $ is connected in series with the combination, then total resistance of combination is $4\Omega $ .

(b) \[9\Omega \]

Ans: If all the three resistance are joined in parallel the resulting resistance will be $3\Omega $ .

7. The  value  of  current  \[I\]  flowing  in  a  given  resistor  for  the  corresponding  values  of potential difference \[V\] across the resistor are given below: 

Plot a graph between \[V\] and \[I\] and calculate the resistance of that resistor. 

Ans: From the given data the \[I - V\] graph is a straight line as shown below: 

Resistance of resistor \[R{\text{ }} = {\text{ }}{V_A} - {V_B}/{1_A} - {1_B}\]

\[ = {\text{ }}12{\text{ }}V{\text{ }}--{\text{ }}6{\text{ }}V/{\text{ }}3.6{\text{ }}A{\text{ }}--{\text{ }}1.8{\text{ }}A\]

\[ = {\text{ }}6V/{\text{ }}1.8{\text{ }}A{\text{ }}\]

\[ = {\text{ }}3.3{\text{ }}\Omega \]

8. Explain the following: 

(a) Why is tungsten used almost exclusively for filament of electric lamps? 

Ans: For the filament of electric lamps, we need a robust metal with a high melting point. Because of its high melting point, tungsten is utilised only for electric lamp filament.

(b) Why  are  the  conductors  of  electric  heating  devices,  such  as  bread-toasters  and electric irons, made of an alloy rather than a pure metal? 

Ans: Electric heating device conductors are composed of an alloy because it has a higher resistance than pure metal and a higher melting point, which prevents it from oxidising at high temperatures.

(c) Why is the series arrangement not used for domestic circuits? 

Ans: As the current to all appliances remains constant despite varying resistance, each appliance cannot be turned on or off independently.

(d) How does the resistance of wire vary with its area of cross-section? 

Ans: As resistance of a wire is inversely proportional to its cross-section area, the resistance will decrease when the area of cross section increases.

(e) Why are copper and aluminum wires usually employed for electric transmission? 

Ans: As copper and aluminium wires are good conductors with low resistance, they are commonly utilised for electrical transmission. They can also be drawn into thin wires since they are ductil

Important Questions of Chapter 11 Class 10 Science - Free PDF Download

Students who are weak in Science and do not have a strong core knowledge might find the chapter electricity quite confusing. This chapter is full of theories and diagrams which confuse students and act as a barrier in their way of achieving good marks. Students must plan to avoid this situation so that they can score the highest possible marks in the final exams. The best way to overcome this problem is a continuous practice. Students can solve some of Class 10 Science Chapter 11 Important Questions regularly. They must induct these practice hours in their preparation schedule. So that every day some time is given to revise and practice. This will make the students efficient and thorough.

Students who are unable to solve the questions must take the help of Important Questions of Electricity Class 10 with Solutions which is available on Vedantu for free. Students can download these questions in PDF format. The downloaded CBSE Class 10 Science Chapter 11 Important Questions act as a guide in preparation for the final exams.

Class 10 Science Ch 11 Important Questions

Students will learn about the concepts and theories of electricity when they will study Chapter 11 of Class 10. They will gain a detailed knowledge of this chapter after practising Important Questions for Class 10 Science Chapter 11. Some of the knowledge that the students will learn are as follows:

Electricity

Electricity is considered as a set of physical phenomena which are associated with the presence and motion of electric charge. It is also believed that electricity is somehow related to magnetism which is why both are part of a phenomenon which is called the phenomenon of electromagnetism. This was described in Maxwell's equations. There are many other phenomena which are related to electricity such as lightning, static electricity, electric heating, electric discharges and many others.

An electric field gets produced, when there is some kind of presence of positive or negative charge. The positive and negative charges are considered as electric charges. When the electric charge moves it is known as Electric current, and it produces a magnetic field nearby.

Attributes of Electricity

There are two primary attributes of electricity: voltage and current. They both have different properties and are quite different from each other, but only in electronic circuits, they are interrelated. Absence of any of these attributes can harm the circuit operation and ruin it. Let's discuss both these attributes:

Voltage:

Voltage is considered as a  force which is responsible for making the current flow in a circuit. Voltage gets measured in volts. In order to get a better understanding of voltage, try to think of a water faucet, in which voltage can get compared to the pressure at which the water is coming out of the faucet. A slow-flowing stream from a faucet resembles a low voltage circuit. In contrast, a high flowing stream resembles a high voltage circuit. A voltage is a mandatory requirement for a current to flow.

Current:

The movement of electrical charges is termed as current. When electrons flow through an electronic circuit, it generates a current. Current is measured in terms of amperes (amp). Here also we can take the example of a water faucet to understand the concept of current. It is assumed that the current is at a high level when more water flows in an hour through the faucet, whereas when the flowing water level is low, the current is also low. After the rain, you must have seen that river flows faster than the usual speed because the current remains high at that point of time as more amount of water passes during rains.

Conductors and Insulators

Another thing that students are going to learn while practising the Electricity Class 10 Important Questions is about the significant classification of elements which is done on the basis of their conductivity of electric charge, i.e. conductors and insulators.

Conductors:

In simple terms, conductors is the name given to those materials which allow electricity to flow through them easily without any difficulties. The conductors' property that makes them capable of allowing electricity to flow through them is termed as conductivity. When electrons start flowing in the conductors, they start to produce an electric current.  The force which is required for the current to move through the conductor is known as the voltage. Some examples of conductors are copper, gold and iron. Let's discuss some properties of a Conductor.

• Materials which are considered as conductors have a minimal resistance because electricity flows through them.

• The inductance of the electric conductor occurs when there is a high voltage drop in the conductor.

• Inside a perfect conductor, the electric field is considered to be zero. It is zero because it helps to keep the electrons calm so that they don’t accelerate.

• There is no electric charge inside the material which is considered as a conductor.

Insulators

Insulators are considered to be materials within which free flow of electrons from one particle of the element to the other particle is interrupted. If a certain amount of charge is transferred to such an element at a given point of time, then the charge does not get distributed in the surface and remains at the same position. The most common process to charge these elements is by rubbing or charging it through induction. Some examples of insulators are wood, plastic and glass. Let's discuss some properties of insulators:

• There are no free electrons in such material because all the electrons are tightly held with each other.

• The ability of these materials to stop the electric current from passing through them is known as resistance.

• Dielectric length of insulators is vast. Dielectric strength is considered as the maximum electric field that an insulator can handle without suffering an electrical breakdown.

• High air permeability is a feature of good insulators as they allow air to pass through their pores.

Important Questions for Electricity Class 10

To give the students an overview of the Important Questions of Chapter 11 Class 10 Science, we are listing here some of the questions which are most likely to come in the exams:

What is Electricity?

• What are the attributes of electricity?

• Write the SI unit of resistivity.

• What do you mean by electric current? Mention and define the SI unit of electric current.

• What do you understand by electrical resistivity of a material? Describe an experiment which will show the factors on which the resistance of a conducting wire depends.

• What do you understand by a conductor? State its properties.

• What do you understand by an insulator? State its properties.

• Differentiate between indicator and conductor.

• State the difference between electric energy and power.

• Mention the commercial unit of electric energy.

• Convert the unit of electric energy into joules.

Benefits of Important Questions for Class 10 Science Chapter 11

• Class 10 Science Chapter 11 Important Questions offer significant advantages for board exam preparation.

• Vedantu's team conducts thorough research to compile a list with a high probability of exam inclusion.

• Expert reviews from seasoned professionals with extensive experience ensure accuracy.

• Questions adhere to the CBSE board format, providing students insight into exam patterns.

• Detailed and explanatory solutions accompany these important questions for comprehensive understanding.

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