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Alternating Current question

2017 · Shift 1 · Q42
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Alternating Current question

2017 · Shift 1 · Q42

JEE AdvancedPhysicsAlternating CurrentMultiple correct+4 / −1
In the circuit shown, L=1 μH,C=1 μF L = 1\,\mu H,C = 1\,\mu F\,L=1μH,C=1μF and R=1 kΩ.R = 1\,k\Omega .R=1kΩ. They are connected in series with an a.c. source V=V0sin⁡ωtV = {V_0}\sin \omega tV=V0​sinωt as shown. Which of the following options is/are correct? JEE Advanced 2017 Paper 1 Offline Physics - Alternating Current Question 15 English
  1. A
    The current will be in phase with the voltage if ω=104rad.s−1\omega = {10^4}rad.{s^{ - 1}}ω=104rad.s−1
  2. B
    The frequency at which the current will be in phase with the voltage is independent of RRR
  3. C
    At ω∼0\omega \sim 0ω∼0 the current flowing through the circuit becomes nearly zero
  4. D
    At ω>>106rad.s−1,\omega \gt \gt {10^6}rad.{s^{ - 1}},ω>>106rad.s−1, the circuit behaves like a capacitor
View written solutionFree

Correct answer: B, C

  1. Series RLC circuit condition for phase

For a series RLCRLCRLC circuit, Z=R+j(ωL−1ωC).Z=R+j\left(\omega L-\frac{1}{\omega C}\right).Z=R+j(ωL−ωC1​).

The current is in phase with the applied voltage when the net reactance is zero: ωL−1ωC=0.\omega L-\frac{1}{\omega C}=0.ωL−ωC1​=0. So, ω2=1LC\omega^2=\frac{1}{LC}ω2=LC1​ and hence the resonant angular frequency is ω0=1LC.\omega_0=\frac{1}{\sqrt{LC}}.ω0​=LC​1​.

Given: L=1 μH=10−6 H,C=1 μF=10−6 FL=1\,\mu H=10^{-6}\,H, \qquad C=1\,\mu F=10^{-6}\,FL=1μH=10−6H,C=1μF=10−6F

Therefore,

=\frac{1}{10^{-6}}=10^6\,\text{rad s}^{-1}.$$ --- 2. **Check option A** Option A says current is in phase with voltage if $$\omega=10^4\,\text{rad s}^{-1}.$$ But we found the in-phase condition at $$\omega_0=10^6\,\text{rad s}^{-1}.$$ So **A is false**. --- 3. **Check option B** The in-phase condition is determined by $$\omega L=\frac{1}{\omega C},$$ which gives $$\omega_0=\frac{1}{\sqrt{LC}}.$$ This expression does **not** contain $R$. So the frequency at which current is in phase with voltage is independent of $R$. Hence **B is true**. --- 4. **Check option C** At very low frequency, $\omega\to 0$: - Inductive reactance: $$X_L=\omega L\to 0$$ - Capacitive reactance: $$X_C=\frac{1}{\omega C}\to \infty$$ So the capacitor offers very large opposition, making the impedance very large: $$|Z|\to \infty.$$ Therefore current $$I=\frac{V_0}{|Z|}\to 0.$$ Thus at $\omega\approx 0$, current becomes nearly zero. So **C is true**. --- 5. **Check option D** At very high frequency, $\omega\gg 10^6\,\text{rad s}^{-1}$: - $$X_L=\omega L$$ becomes very large - $$X_C=\frac{1}{\omega C}$$ becomes very small Hence net reactance $$X=\omega L-\frac{1}{\omega C}\approx \omega L>0,$$ so the circuit behaves like an **inductor**, not a capacitor. Therefore **D is false**. --- 6. **Final answer** Correct options are: $$\boxed{B,\ C}$$ This matches the stored correct answer.
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