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Alternating Current question

2006 · Shift 0 · Q94
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Alternating Current question

2006 · Shift 0 · Q94

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
In a series resonant LCRLCRLCR circuit, the voltage across RRR is 100100100 volts and R=1 kΩR = 1\,k\OmegaR=1kΩ with C=2μF.C = 2\mu F.C=2μF. The resonant frequency ω\omegaω is 200rad/s200rad/s200rad/s. At resonance the voltage across LLL is
  1. A
    2.5×10−2V2.5 \times {10^{ - 2}}V2.5×10−2V
  2. B
    40V40V40V
  3. C
    250V250V250V
  4. D
    4×10−3V4 \times {10^{ - 3}}V4×10−3V
View written solutionFree

Correct answer: C

  1. Use resonance condition in a series LCRLCRLCR circuit

At resonance,

XL=XCX_L = X_CXL​=XC​

and the current is maximum.

Also, the voltage across the resistor is

VR=IRV_R = IRVR​=IR

Given:

VR=100 V,R=1 kΩ=1000 ΩV_R = 100\text{ V}, \quad R = 1\,k\Omega = 1000\,\OmegaVR​=100 V,R=1kΩ=1000Ω

So the current is

I=VRR=1001000=0.1 AI = \frac{V_R}{R} = \frac{100}{1000} = 0.1\text{ A}I=RVR​​=1000100​=0.1 A
  1. Find capacitive reactance at resonance

Given:

C=2μF=2×10−6F,ω=200 rad/sC = 2\mu F = 2\times 10^{-6} F, \quad \omega = 200\text{ rad/s}C=2μF=2×10−6F,ω=200 rad/s

Capacitive reactance:

XC=1ωCX_C = \frac{1}{\omega C}XC​=ωC1​

Substitute values:

XC=1200×2×10−6=14×10−4=2500 ΩX_C = \frac{1}{200 \times 2\times 10^{-6}} = \frac{1}{4\times 10^{-4}} = 2500\,\OmegaXC​=200×2×10−61​=4×10−41​=2500Ω

Since at resonance,

XL=XC=2500 ΩX_L = X_C = 2500\,\OmegaXL​=XC​=2500Ω
  1. Find voltage across inductor

Voltage across inductor:

VL=IXLV_L = IX_LVL​=IXL​

So,

VL=0.1×2500=250 VV_L = 0.1 \times 2500 = 250\text{ V}VL​=0.1×2500=250 V
  1. Match with options

Thus,

VL=250 VV_L = 250\text{ V}VL​=250 V

So the correct option is:

C: 250 V250\text{ V}250 V

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