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An ac voltage is applied to a resistance R and an inductor L in series. If R and the inductive reactance are both equal to 30 Ω, the phase difference between the applied voltage and the current in the circuit is

1. π/6
2. π/4
3. π/2
4. zero

Correct answer: π/4

Solution

The phase difference in an RL circuit is given by \( \phi = \tan^{-1}(X_L / R) \), where \( X_L \) is the inductive reactance. Since \( X_L = R = 30 \ \Omega \), \( \phi = \tan^{-1}(30/30) = \tan^{-1}(1) = \pi/4 \).

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