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JEE PHYSICS MCQ: ALTERNATING CURRENT

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An inductor and a resistor are connected in series with an ac source. In this circuit.
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Correct option: B
The current and the PD across the resistance lag behind the PD across the inductance by an angle π/2\pi / 2
The correct answer is B (The current and the PD across the resistance lag behind the PD across the inductance by an angle π/2\pi / 2) because in an inductor, voltage leads current by 90 degrees, and in a resistor, voltage and current are in phase.
Solution approach
  1. Correct Answer + Core Reason: The current and the PD across the resistance lag behind the PD across the inductance by an angle π/2\pi / 2. In an RL series circuit, the voltage across the inductor leads the current by 90 degrees, while the voltage and current across the resistor are in phase.

    Step-by-Step Solution:

    Step 1: In a resistor, the voltage and current are in phase.

    Step 2: In an inductor, the voltage leads the current by π/2\pi/2 (90 degrees).

    Step 3: In a series RL circuit, the same current flows through both the resistor and the inductor.

    Step 4: Since the voltage across the inductor leads the current by π/2\pi/2, and the voltage across the resistor is in phase with the current, the voltage across the resistor lags behind the voltage across the inductor by π/2\pi/2.

    Why other options are wrong:
    - Option A: The current and the PD across the resistance lead the PD across the inductance is incorrect because the voltage across the inductor leads the current.
    - Option C: The current and the PD across the resistance lag behind the PD across the inductance by an angle π\pi is incorrect because the phase difference is π/2\pi/2.
    - Option D: The PD across the resistance lags behind the PD across the inductance by an angle π/2\pi / 2 but the current in resistance leads the PD across the inductance by π/2\pi / 2 is incorrect because the current in the resistance is in phase with the voltage across the resistance.

    Key Concepts:
    - AC Circuits
    - Resistors
    - Inductors
    - Phase Difference
    - Voltage and Current Relationships

    Memory Aids:
    - ELI the ICE man: E leads I in an inductor (L), I leads E in a capacitor (C).

    Quick Tips:
    - Remember the phase relationships between voltage and current in resistors and inductors.

More Questions from ALTERNATING CURRENT

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In a series RC circuit with an AC source, R=300 ΩR=300 \, \Omega, C=25 μF,ε0=50 VC=25 \, \mu \mathrm{F}, \varepsilon_0=50 \mathrm{~V} and v=50/π Hzv=50 / \pi \, \mathrm{Hz}. Find the peak current and the average power dissipated in the circuit. In a series circuit R=300Ω,L=0.9 H,C=2.0R=300 \Omega, L=0.9 \mathrm{~H}, C=2.0 μF\mu F and ω=1000 rad/sec\omega=1000 \mathrm{~rad} / \mathrm{sec}. The impedance of the circuit is: An L-C-R series circuit with a resistance of 100Ω100 \Omega is connected to 200 V200 \mathrm{~V} (AC source) and angular frequency 300 rad/s300 \, \mathrm{rad} / \mathrm{s}. When only the capacitor is removed, then the current lags behind the voltage by 60∘60^{\circ}. When only the inductor is removed the current leads the voltage by 60∘60^{\circ}. The average power dissipated in original L-C-R circuit is Assertion (A) : In series L−C⋅R,ACL-C \cdot R, \mathrm{AC} circuit, current and voltage are in same phase at resonance.

Reason (R) : In series L−C−R,ACL-C-R, \mathrm{AC} circuit, resonant frequency does not depend on the value of resistance. Hence, current at resonance does not depend on resistance.
A series L−C−RL-C-R circuit is connected to an AC\mathrm{AC} voltage source. When LL is removed from the circuit, the phase difference between current and voltage is π4\frac{\pi}{4}. If instead CC is removed from the circuit, the phase difference is again π4\frac{\pi}{4} between current and voltage. The power factor of the circuit is A capacitor in an ideal LC circuit is fully charged by a DC source. Then it is disconnected from DC source, the current in the circuit
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