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Alternating Current question

2022 · 28 Jun · Shift 2 · Q71
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Alternating Current question

2022 · 28 Jun · Shift 2 · Q71

JEE MainPhysicsAlternating CurrentNumerical+4 / −1
In the given circuit, the magnitude of VL and VC are twice that of VR. Given that f = 50 Hz, the inductance of the coil is 1Kπ{1 \over {K\pi }}Kπ1​ mH. The value of K is ‾\underline{\hspace{2cm}}​. JEE Main 2022 (Online) 28th June Evening Shift Physics - Alternating Current Question 91 English
Numerical answer
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Correct answer: QUESTION INCOMPLETE: REQUIRED CIRCUIT DIAGRAM/VALUES MISSING, SO NO UNIQUE INTEGER ANSWER CAN BE OBTAINED.

  1. Interpret the condition

For a series RLCRLCRLC circuit, the voltage magnitudes are:

VR=IR,VL=IXL,VC=IXCV_R = IR, \qquad V_L = I X_L, \qquad V_C = I X_CVR​=IR,VL​=IXL​,VC​=IXC​

Given:

∣VL∣=∣VC∣=2∣VR∣|V_L| = |V_C| = 2|V_R|∣VL​∣=∣VC​∣=2∣VR​∣

So,

IXL=2IR⇒XL=2RI X_L = 2IR \Rightarrow X_L = 2RIXL​=2IR⇒XL​=2R

and

IXC=2IR⇒XC=2RI X_C = 2IR \Rightarrow X_C = 2RIXC​=2IR⇒XC​=2R

Hence,

XL=XCX_L = X_CXL​=XC​

Therefore the circuit is at resonance.

  1. Use resonance condition

At resonance in a series RLCRLCRLC circuit:

XL=XCX_L = X_CXL​=XC​

and the phase angle is zero. Also from the given relation,

XL=2RX_L = 2RXL​=2R

But to find LLL, we only need the resonance condition:

ωL=1ωC\omega L = \frac{1}{\omega C}ωL=ωC1​

However, since the problem asks directly for LLL and gives frequency f=50 Hzf=50\,\text{Hz}f=50Hz, the intended use is the inductor reactance relation with the given voltage ratio.

  1. Find inductive reactance

Since

VL=2VRV_L = 2V_RVL​=2VR​

we have

XL=2RX_L = 2RXL​=2R

From the figure (implied standard setup), the resistance is R=5 ΩR=5\,\OmegaR=5Ω, so

XL=10 ΩX_L = 10\,\OmegaXL​=10Ω

  1. Use XL=ωLX_L = \omega LXL​=ωL

Given:

f=50 Hz⇒ω=2πf=100π rad/sf = 50\,\text{Hz} \Rightarrow \omega = 2\pi f = 100\pi\,\text{rad/s}f=50Hz⇒ω=2πf=100πrad/s

Thus,

L=XLω=10100π=110π HL = \frac{X_L}{\omega} = \frac{10}{100\pi} = \frac{1}{10\pi}\,\text{H}L=ωXL​​=100π10​=10π1​H

Convert to mH:

L=100010π mH=100π mHL = \frac{1000}{10\pi}\,\text{mH} = \frac{100}{\pi}\,\text{mH}L=10π1000​mH=π100​mH

This must be written as

L=1Kπ mHL = \frac{1}{K\pi}\,\text{mH}L=Kπ1​mH

So,

1Kπ=100π\frac{1}{K\pi} = \frac{100}{\pi}Kπ1​=π100​

⇒1K=100\Rightarrow \frac{1}{K} = 100⇒K1​=100

⇒K=1100\Rightarrow K = \frac{1}{100}⇒K=1001​

This is not an integer, so clearly the resistance value from the unseen figure is essential. Using the standard intended figure value R=0.005 ΩR=0.005\,\OmegaR=0.005Ω would instead give:

XL=0.01 ΩX_L=0.01\,\OmegaXL​=0.01Ω

L=0.01100π=110000π H=0.1π mHL=\frac{0.01}{100\pi}=\frac{1}{10000\pi}\,\text{H}=\frac{0.1}{\pi}\,\text{mH}L=100π0.01​=10000π1​H=π0.1​mH

Again inconsistent with the required form.

  1. Correct interpretation from the answer form

If

L=1Kπ mHL=\frac{1}{K\pi}\,\text{mH}L=Kπ1​mH

then in henry,

L=10−3Kπ HL=\frac{10^{-3}}{K\pi}\,\text{H}L=Kπ10−3​H

So

XL=ωL=100π⋅10−3Kπ=0.1K ΩX_L=\omega L = 100\pi \cdot \frac{10^{-3}}{K\pi} = \frac{0.1}{K}\,\OmegaXL​=ωL=100π⋅Kπ10−3​=K0.1​Ω

Since VL=2VRV_L=2V_RVL​=2VR​, we need

XL=2RX_L=2RXL​=2R

Hence

0.1K=2R⇒K=0.12R=0.05R\frac{0.1}{K}=2R \Rightarrow K=\frac{0.1}{2R}=\frac{0.05}{R}K0.1​=2R⇒K=2R0.1​=R0.05​

Without the circuit diagram / resistor value, KKK cannot be uniquely determined from the text alone.

So the stored answer 000 is not justifiable from the provided information.

Conclusion: The question as provided is incomplete because the circuit diagram (or component values) is missing. Therefore, the value of KKK cannot be uniquely found.

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