
If 1 unit of electricity costs $0.20$ , how much does it cost to switch on a heater marked 120V, 3A for 90 minutes ?
A) 0.11
B) 2.70
C) 64.80
D) 180.00
Answer
131.7k+ views
Hint: The cost of an electronic Appliance depends on the amount of energy used by it. The commercial unit of energy is kWh. Thus the cost to switch on the heater can be calculated if we know the value of energy used by it in commercial units, which is kWh. Then we can find the cost by multiplying the cost of 1 electricity unit and the total kWh of energy consumed.
Formula used:
$Energy = Power \times Time$
$Power = Voltage \times Current$
Complete step by step solution:
Let V be the voltage across the heater
Voltage across the heater $V = 120$ volts
Converting it into kilovolts (therefore dividing it by 1000, as $1kV = 1000$ volts)
Thus, $V=0.12kV$
Let I be the current in the heater
Current flowing through the heater $I = 3A$
Formula for power of the heater is given by $P = VI$
By substituting the values, we get the value of power of the heater,
\[\therefore \;P = 0.12 \times 3 = 0.36\;kW\]
Let T be the time for which heater is used
Time of usage $(T) = 90{\text{ minutes}}$
Converting it into hours (therefore dividing it by $60$ , as \[1hr = 60mins\] )
Thus time of usage of heater\[ = 1.5{\text{ }}hrs\]
Thus energy consumed is given by $E = PT$
Substituting values,
\[E = 0.36 \times 1.5 = 0.54\;kWh\]
Now the total electricity units consumed by the heater are \[0.54\] kWh.
And cost of 1 unit of electricity \[ = 0.20\]
Cost to switch on heater for 90 minutes \[ = {\text{ }}\;0.54 \times 0.2\] \[ = 0.11\]
Hence, the right answer is (A) that is \[0.11\]
Note: The cost of using any electronic appliance can also be found by using a shortcut formula, that is
Cost \[ = {\text{ }}\dfrac{{\left( {V \times I \times T \times C} \right)}}{{1000}}\] where $V$ voltage in volts, $I$ is current in ampere, $T$ is time in hours and $C$ is cost of one electricity unit.
Formula used:
$Energy = Power \times Time$
$Power = Voltage \times Current$
Complete step by step solution:
Let V be the voltage across the heater
Voltage across the heater $V = 120$ volts
Converting it into kilovolts (therefore dividing it by 1000, as $1kV = 1000$ volts)
Thus, $V=0.12kV$
Let I be the current in the heater
Current flowing through the heater $I = 3A$
Formula for power of the heater is given by $P = VI$
By substituting the values, we get the value of power of the heater,
\[\therefore \;P = 0.12 \times 3 = 0.36\;kW\]
Let T be the time for which heater is used
Time of usage $(T) = 90{\text{ minutes}}$
Converting it into hours (therefore dividing it by $60$ , as \[1hr = 60mins\] )
Thus time of usage of heater\[ = 1.5{\text{ }}hrs\]
Thus energy consumed is given by $E = PT$
Substituting values,
\[E = 0.36 \times 1.5 = 0.54\;kWh\]
Now the total electricity units consumed by the heater are \[0.54\] kWh.
And cost of 1 unit of electricity \[ = 0.20\]
Cost to switch on heater for 90 minutes \[ = {\text{ }}\;0.54 \times 0.2\] \[ = 0.11\]
Hence, the right answer is (A) that is \[0.11\]
Note: The cost of using any electronic appliance can also be found by using a shortcut formula, that is
Cost \[ = {\text{ }}\dfrac{{\left( {V \times I \times T \times C} \right)}}{{1000}}\] where $V$ voltage in volts, $I$ is current in ampere, $T$ is time in hours and $C$ is cost of one electricity unit.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

Degree of Dissociation and Its Formula With Solved Example for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Diffraction of Light - Young’s Single Slit Experiment

Clemmenson and Wolff Kishner Reductions for JEE

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation
