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Alternating Current question

2023 · Shift 1 · Q50
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Alternating Current question

2023 · Shift 1 · Q50

JEE AdvancedPhysicsAlternating CurrentMCQ+3 / −1
A series LCR circuit is connected to a 45sin⁡(ωt)45 \sin (\omega t)45sin(ωt) Volt source. The resonant angular frequency of the circuit is 105 rad s−110^5 ~\mathrm{rad}~ \mathrm{s}^{-1}105 rad s−1 and current amplitude at resonance is I0I_0I0​. When the angular frequency of the source is ω=8×104 rad s−1\omega=8 \times 10^4 ~\mathrm{rad} ~\mathrm{s}^{-1}ω=8×104 rad s−1, the current amplitude in the circuit is 0.05I00.05 I_00.05I0​. If L=50 mHL=50 ~\mathrm{mH}L=50 mH, match each entry in List-I with an appropriate value from List-II and choose the correct option.

List - I List - II
(P) I0I_0I0​ in mA\mathrm{mA}mA (1) 44.4
(Q) The quality factor of the circuit (2) 18
(R) The bandwidth of the circuit in rad s−1\mathrm{rad}~ \mathrm{s}^{-1}rad s−1 (3) 400
(S) The peak power dissipated at resonance in Watt (4) 2250
(5) 500
  1. A
    P→2,Q→3,R→5,S→1P \rightarrow 2, Q \rightarrow 3, R \rightarrow 5, S \rightarrow 1P→2,Q→3,R→5,S→1
  2. B
    P→3,Q→1,R→4,S→2P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 2P→3,Q→1,R→4,S→2
  3. C
    P→4,Q→5,R→3,S→1P \rightarrow 4, Q \rightarrow 5, R \rightarrow 3, S \rightarrow 1P→4,Q→5,R→3,S→1
  4. D
    P→4,Q→2,R→1,S→5P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 5P→4,Q→2,R→1,S→5
View written solutionFree

Correct answer: B

  1. Given data
  • Source voltage: v=45sin⁡(ωt)v = 45\sin(\omega t)v=45sin(ωt) V, so voltage amplitude is V0=45 VV_0 = 45\ \text{V}V0​=45 V
  • Resonant angular frequency: ω0=105 rad s−1\omega_0 = 10^5\ \text{rad s}^{-1}ω0​=105 rad s−1
  • Inductance: L=50 mH=0.05 HL = 50\ \text{mH} = 0.05\ \text{H}L=50 mH=0.05 H
  • At resonance, current amplitude is I0I_0I0​.
  • At ω=8×104 rad s−1\omega = 8\times 10^4\ \text{rad s}^{-1}ω=8×104 rad s−1 current amplitude is I=0.05I0I = 0.05 I_0I=0.05I0​
  1. Use current ratio to find resistance

For a series LCR circuit, I=V0Z,I0=V0RI = \frac{V_0}{Z}, \qquad I_0 = \frac{V_0}{R}I=ZV0​​,I0​=RV0​​ At resonance, impedance is minimum and equals RRR.

Given: II0=0.05\frac{I}{I_0} = 0.05I0​I​=0.05 So, V0/ZV0/R=0.05\frac{V_0/Z}{V_0/R} = 0.05V0​/RV0​/Z​=0.05 RZ=0.05\frac{R}{Z} = 0.05ZR​=0.05 Z=20RZ = 20RZ=20R

Now, Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}Z=R2+(XL​−XC​)2​ Thus, R2+(XL−XC)2=20R\sqrt{R^2 + (X_L - X_C)^2} = 20RR2+(XL​−XC​)2​=20R R2+(XL−XC)2=400R2R^2 + (X_L - X_C)^2 = 400R^2R2+(XL​−XC​)2=400R2 (XL−XC)2=399R2 (X_L - X_C)^2 = 399R^2(XL​−XC​)2=399R2 ∣XL−XC∣=399 R|X_L - X_C| = \sqrt{399}\,R∣XL​−XC​∣=399​R

  1. Compute XL−XCX_L - X_CXL​−XC​ at ω=8×104\omega = 8\times 10^4ω=8×104 rad/s

First find CCC using resonance condition: ω0=1LC\omega_0 = \frac{1}{\sqrt{LC}}ω0​=LC​1​ So, C=1Lω02C = \frac{1}{L\omega_0^2}C=Lω02​1​ C=10.05×(105)2=15×108=2×10−9 FC = \frac{1}{0.05\times (10^5)^2} = \frac{1}{5\times 10^8} = 2\times 10^{-9}\ \text{F}C=0.05×(105)21​=5×1081​=2×10−9 F

Now at ω=8×104\omega = 8\times 10^4ω=8×104: XL=ωL=8×104×0.05=4000 ΩX_L = \omega L = 8\times 10^4 \times 0.05 = 4000\ \OmegaXL​=ωL=8×104×0.05=4000 Ω

XC=1ωC=1(8×104)(2×10−9)=11.6×10−4=6250 ΩX_C = \frac{1}{\omega C} = \frac{1}{(8\times 10^4)(2\times 10^{-9})} = \frac{1}{1.6\times 10^{-4}} = 6250\ \OmegaXC​=ωC1​=(8×104)(2×10−9)1​=1.6×10−41​=6250 Ω

Hence, ∣XL−XC∣=∣4000−6250∣=2250 Ω|X_L - X_C| = |4000 - 6250| = 2250\ \Omega∣XL​−XC​∣=∣4000−6250∣=2250 Ω

So, 2250=399 R2250 = \sqrt{399}\,R2250=399​R R≈225020=112.5 ΩR \approx \frac{2250}{20} = 112.5\ \OmegaR≈202250​=112.5 Ω

(Using 399≈20\sqrt{399} \approx 20399​≈20 for matching values.)

  1. Find I0I_0I0​

At resonance, I0=V0R=45112.5=0.4 A=400 mAI_0 = \frac{V_0}{R} = \frac{45}{112.5} = 0.4\ \text{A} = 400\ \text{mA}I0​=RV0​​=112.545​=0.4 A=400 mA

So, P→400P \rightarrow 400P→400 Thus, P→(3)P \rightarrow (3)P→(3)

  1. Find quality factor QQQ

For series LCR circuit, Q=ω0LRQ = \frac{\omega_0 L}{R}Q=Rω0​L​ Q=105×0.05112.5=5000112.5=44.4Q = \frac{10^5\times 0.05}{112.5} = \frac{5000}{112.5} = 44.4Q=112.5105×0.05​=112.55000​=44.4

So, Q→44.4Q \rightarrow 44.4Q→44.4 Thus, Q→(1)Q \rightarrow (1)Q→(1)

  1. Find bandwidth

Bandwidth for a series resonant circuit is Δω=ω0Q=RL\Delta \omega = \frac{\omega_0}{Q} = \frac{R}{L}Δω=Qω0​​=LR​ Using R/LR/LR/L: Δω=112.50.05=2250 rad s−1\Delta \omega = \frac{112.5}{0.05} = 2250\ \text{rad s}^{-1}Δω=0.05112.5​=2250 rad s−1

So, R→2250R \rightarrow 2250R→2250 Thus, R→(4)R \rightarrow (4)R→(4)

  1. Find peak power dissipated at resonance

At resonance, power factor is 1, and instantaneous peak power is pmax⁡=V0I0=45×0.4=18 Wp_{\max} = V_0 I_0 = 45\times 0.4 = 18\ \text{W}pmax​=V0​I0​=45×0.4=18 W

So, S→18S \rightarrow 18S→18 Thus, S→(2)S \rightarrow (2)S→(2)

  1. Final matching
  • P→3P \rightarrow 3P→3
  • Q→1Q \rightarrow 1Q→1
  • R→4R \rightarrow 4R→4
  • S→2S \rightarrow 2S→2

This corresponds to Option B.

  1. Comparison with stored answer

Stored correct answer: B

Derived answer: B

So the derived answer agrees with the stored answer.

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