An inductor and resistor are connected in series with an AC source. In this circuit,
The current and the PD across resistance lead the PD across the inductance.
(A) The current and the PD across the resistance lag behind the PD across the inductance by angle $\dfrac{\pi }{2}$
(B) The current and the PD across the resistance lag behind the PD across the inductance by an angle $\pi $
(C) The PD across the resistance lags behind the PD across the inductance by an angle (D)$\dfrac{\pi }{2}$ but the current in resistance leads the PD across the inductance by $\dfrac{\pi }{2}$.
Answer
Verified
118.8k+ 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
How to find Oxidation Number - Important Concepts for JEE
How Electromagnetic Waves are Formed - Important Concepts for JEE
Electrical Resistance - Important Concepts and Tips for JEE
Average Atomic Mass - Important Concepts and Tips for JEE
Chemical Equation - Important Concepts and Tips for JEE
Concept of CP and CV of Gas - Important Concepts and Tips for JEE
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Chemistry Question Paper with Answer Keys and Solutions
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
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 (Free PDF Download)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
Degree of Dissociation and Its Formula With Solved Example for JEE
Diffraction of Light - Young’s Single Slit Experiment
JEE Main 2025: Derivation of Equation of Trajectory in Physics