JEE Challenger
More from Alternating Current
PhyMedAdv 2017 • Paper 1

Analysis of AC Circuit with Resistor Inductor and Capacitor

Consider the circuit shown where L=1 μHL = 1\ \mu\text{H}, C=1 μFC = 1\ \mu\text{F}, and R=1 kΩR = 1\ \text{k}\Omega are connected in series with an alternating source V=V0sin⁡ωtV = V_0 \sin \omega t. Identify the correct option(s) from the following:

Question Diagram 1

Options

A

The current will be in phase with the voltage if ω=104 rad⋅s−1\omega = 10^4\ \text{rad}\cdot\text{s}^{-1}

B

The frequency at which the current will be in phase with the voltage is independent of RR

Correct
C

At ω∼0\omega \sim 0 the current flowing through the circuit becomes nearly zero

Correct
D

At ω≫106 rad⋅s−1\omega \gg 10^6\ \text{rad}\cdot\text{s}^{-1}, the circuit behaves like a capacitor

Topics & Concepts

Alternating CurrentAC Circuits

Step-by-Step Solution

To determine the correct options, let us analyze the series LCRLCR circuit driven by an alternating voltage source V=V0sin⁡(ωt)V = V_0 \sin(\omega t).

1. Resonance Frequency and Phase Condition

For the current to be in phase with the applied voltage, the net reactance of the circuit must be zero. This occurs at the resonant angular frequency ω0\omega_0: XL=XC  ⟹  ω0L=1ω0CX_L = X_C \implies \omega_0 L = \frac{1}{\omega_0 C}

Solving for ω0\omega_0: ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}

Given values:

  • L=1 μH=10−6 HL = 1\ \mu\text{H} = 10^{-6}\ \text{H}
  • C=1 μF=10−6 FC = 1\ \mu\text{F} = 10^{-6}\ \text{F}
  • R=1 kΩ=103 ΩR = 1\ \text{k}\Omega = 10^3\ \Omega

Substituting these values into the resonance equation: ω0=110−6×10−6=110−6=106 rad⋅s−1\omega_0 = \frac{1}{\sqrt{10^{-6} \times 10^{-6}}} = \frac{1}{10^{-6}} = 10^6\ \text{rad}\cdot\text{s}^{-1}

  • Option A: Claims current is in phase with voltage at ω=104 rad⋅s−1\omega = 10^4\ \text{rad}\cdot\text{s}^{-1}. Since ω0=106 rad⋅s−1\omega_0 = 10^6\ \text{rad}\cdot\text{s}^{-1}, Option A is incorrect.
  • Option B: The expression ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}} is entirely independent of the resistance RR. Therefore, Option B is correct.

2. Behavior at Low Frequencies (ω∼0\omega \sim 0)

The total impedance ZZ of a series LCRLCR circuit is given by: Z=R2+(ωL−1ωC)2Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}

As ω→0\omega \to 0: XC=1ωC→∞X_C = \frac{1}{\omega C} \to \infty Z→∞Z \to \infty

The peak current amplitude is: I0=V0ZI_0 = \frac{V_0}{Z}

Since Z→∞Z \to \infty, the current amplitude I0→0I_0 \to 0. Thus, at frequencies close to zero, the current flowing through the circuit becomes nearly zero. Therefore, Option C is correct.

3. Behavior at High Frequencies (ω≫106 rad⋅s−1\omega \gg 10^6\ \text{rad}\cdot\text{s}^{-1})

At angular frequencies much larger than the resonant frequency (ω≫ω0=106 rad⋅s−1\omega \gg \omega_0 = 10^6\ \text{rad}\cdot\text{s}^{-1}):

  • Inductive reactance: XL=ωL≫ω0L=1 ΩX_L = \omega L \gg \omega_0 L = 1\ \Omega
  • Capacitive reactance: XC=1ωC≪1ω0C=1 ΩX_C = \frac{1}{\omega C} \ll \frac{1}{\omega_0 C} = 1\ \Omega

Since XL≫XCX_L \gg X_C, the inductive reactance dominates over the capacitive reactance. Hence, the circuit behaves like an inductive circuit (inductor), not a capacitor. Therefore, Option D is incorrect.

Conclusion

The correct options are B and C.

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