Answer
Verified
114.6k+ views
Hint Find the impedance for series resistance and inductor AC circuit then, draw the phasor diagram of series inductance and resistance series AC circuit. In the phasor diagram, draw the voltage across resistance and also draw the voltage across inductance. If the angle of voltage across inductance is greater then it leads the voltage across resistance.
Step by Step Solution
Let the resistance and inductance be R and L respectively
Construct a series L-R AC circuit
Now, we get
${V_{net}} = {\vec V_L} + {\vec V_R}$
By taking magnitude
\[
\left| {{{\vec V}_{net}}} \right| = \sqrt {({V^2}_L + {V^2}_R)} \\
\left| {{{\vec V}_{net}}} \right| = \sqrt {({i^2}{X^2}_L + {i^2}{R^2})} \\
\\
\]
By taking ${i^2}$ common
\[\left| {{{\vec V}_{net}}} \right| = i\sqrt {({X^2}_L + {R^2})} \]
To calculate the impedance let impedance be $Z$
$Z = \dfrac{V}{i} = \sqrt {({X_L}^2 + {R^2})} $
Because, ${X_L} = \omega L$ we get
$Z = \sqrt {({\omega ^2}{L^2} + {R^2})} $
This is the impedance for series L-R AC circuit
Now, draw the phasor diagram for series L-R AC circuit
$
\tan \varphi = \dfrac{{{V_L}}}{{{V_R}}} \\
\tan \varphi = \dfrac{{i{X_L}}}{{iR}} \\
\\
$
Now, we get
$\varphi = {\tan ^{ - 1}}\left( {\dfrac{{{X_L}}}{R}} \right)$
Therefore, after seeing the phasor diagram we can easily conclude that potential difference across inductance leads the current and potential difference across resistance by an angle $\dfrac{\pi }{2}$
Hence, option (B) is the correct answer.
Note resistance is the measure of the opposition of current in the circuit. It is denoted by R. It is measured in ohms $(\Omega )$.
Inductance is the property of a conductor to oppose the flow of electric current through it. The flow of current makes a magnetic field around the conductor. It is denoted by L. It is measured in Henry $(H)$.
Step by Step Solution
Let the resistance and inductance be R and L respectively
Construct a series L-R AC circuit
Now, we get
${V_{net}} = {\vec V_L} + {\vec V_R}$
By taking magnitude
\[
\left| {{{\vec V}_{net}}} \right| = \sqrt {({V^2}_L + {V^2}_R)} \\
\left| {{{\vec V}_{net}}} \right| = \sqrt {({i^2}{X^2}_L + {i^2}{R^2})} \\
\\
\]
By taking ${i^2}$ common
\[\left| {{{\vec V}_{net}}} \right| = i\sqrt {({X^2}_L + {R^2})} \]
To calculate the impedance let impedance be $Z$
$Z = \dfrac{V}{i} = \sqrt {({X_L}^2 + {R^2})} $
Because, ${X_L} = \omega L$ we get
$Z = \sqrt {({\omega ^2}{L^2} + {R^2})} $
This is the impedance for series L-R AC circuit
Now, draw the phasor diagram for series L-R AC circuit
$
\tan \varphi = \dfrac{{{V_L}}}{{{V_R}}} \\
\tan \varphi = \dfrac{{i{X_L}}}{{iR}} \\
\\
$
Now, we get
$\varphi = {\tan ^{ - 1}}\left( {\dfrac{{{X_L}}}{R}} \right)$
Therefore, after seeing the phasor diagram we can easily conclude that potential difference across inductance leads the current and potential difference across resistance by an angle $\dfrac{\pi }{2}$
Hence, option (B) is the correct answer.
Note resistance is the measure of the opposition of current in the circuit. It is denoted by R. It is measured in ohms $(\Omega )$.
Inductance is the property of a conductor to oppose the flow of electric current through it. The flow of current makes a magnetic field around the conductor. It is denoted by L. It is measured in Henry $(H)$.
Recently Updated Pages
JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key
JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key
Hybridization of Atomic Orbitals Important Concepts and Tips for JEE
Atomic Structure: Complete Explanation for JEE Main 2025
Trending doubts
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
Collision - Important Concepts and Tips for JEE
Ideal and Non-Ideal Solutions Raoult's Law - JEE
Current Loop as Magnetic Dipole and Its Derivation for JEE
Two plane mirrors are inclined at angle theta as shown class 12 physics JEE_Main