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In the AC circuit shown, an ideal source feeds two parallel branches: branch 1 contains a resistor R in series with an inductor of reactance xL, and branch 2 contains a pure capacitor of reactance xC. If I1 is the current in the R-L branch and I2 is the current in the capacitor branch, the phase difference between I1 and I2 is

1. pi/2 + tan⁻¹(xL/R)
2. pi/2 - tan⁻¹(xL/R)
3. tan⁻¹((xL - xC)/R)
4. tan⁻¹((xL - xC)/R) + pi/2

Correct answer: pi/2 + tan⁻¹(xL/R)

Solution

In a parallel combination the source voltage V is common to both branches, so the phase of each branch current is measured against the same reference V. In branch 1 (R in series with L), the current I1 lags V by an angle phi1 = tan⁻¹(xL/R). In branch 2 (pure capacitor), the current I2 leads V by phi2 = pi/2. Since I1 is on the lagging side and I2 on the leading side of V, the total angle between them is the sum: phi = phi2 + phi1 = pi/2 + tan⁻¹(xL/R). The options tan⁻¹((xL - xC)/R) describe a single series R-L-C branch, which does not apply to two separate parallel branches, so they are distractors.

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