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In an LCR series AC circuit, the voltage across the resistor is 2 V. The inductive reactance XL = 10 ohm and the capacitive reactance XC = 5 ohm. The phase difference between the voltage across the circuit and the current is 30 deg. Match each quantity in Column I with its value in Column II. Column I: I. Resistance R (in ohm) II. Current I (in A) III. Voltage of AC source (in V) IV. Impedance Z (in ohm) Column II: P. 4/sqrt(3) Q. 10 R. 5*sqrt(3) S. 2/(5*sqrt(3))

1. I – Q; II – R; III – S; IV – P
2. I – R; II – P; III – S; IV – Q
3. I – R; II – S; III – P; IV – Q
4. I – S; II – Q; III – P; IV – R

Correct answer: I – R; II – S; III – P; IV – Q

Solution

From tan(phi) = (XL-XC)/R: R = 5/tan(30 deg) = 5*sqrt(3) ohm. I = VR/R = 2/(5*sqrt(3)) A. Z = R/cos(30 deg) = 5*sqrt(3)/(sqrt(3)/2) = 10 ohm. V_source = I*Z = 2/(5*sqrt(3)) * 10 = 4/sqrt(3) V.

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