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For a given L - C - R circuit, list 1 - gives voltage across source inductor, capacitor & resistance, and list -2 gives their magnitudes.
List -1List -II(I)(p)480(II)VL(Q)400(III)VC(R)240(IV)VR(S)200(T)160(U)0
If current in the circuit at an instant is 6A, then correct match for potential difference at that instant. [Neglect negative sign if any]

A
IP IIS IIIT IVU
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B
IP IIQ IIIT IVR
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C
IQ IIS IIIT IVR
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D
IS IIT IIIR IVU
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Solution

The correct option is B IP IIQ IIIT IVR
IP, IIQ, IIIT, IVR
For given circuit = R=40Ω,XC=20Ω&XL=50Ω
& Z=R2+(XLXC)2=50Ω
i0=0z=50050=10A

...
tanϕ=3040ϕ=37
...
VoL=i0×XL=10×50=500
VoC=i0×XC=10×20=200

Say i=i0sin θ, then VL=VoL cos θ and VR=VoRsin θ
Given,
i=6=10sin θθ=37
VL=VoL cos θ=500×45=400V
VC=VoCcos θ=200×45=160V
VR=VoRsin θ=240V

Since, current lags voltage by ϕ, voltage is given by,
=0 sin (θ+ϕ)=480 V
where VoL=i0XL=500V

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