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A resistance R in series with circuit element A causes the current to lag the voltage by angle theta1. The same R in series with element B causes the current to lead the voltage by angle theta2. When R, A and B are all connected in series with the same AC source, the current leads the voltage by angle theta. Which relation holds?
- theta = theta2 - theta1
- tan(theta) = tan(theta2) - tan(theta1)
- theta = (theta1 + theta2)/2
- None of these
Correct answer: None of these
Solution
With R+A: tan(theta1) = X_A/R (A inductive). With R+B: tan(theta2) = X_B/R (B capacitive, leading). With R+A+B in series, total resistance is still R (A and B are reactive), and net reactance is X_B - X_A, so tan(theta) = (X_B - X_A)/R = X_B/R - X_A/R = tan(theta2) - tan(theta1). This would suggest option 2 - but only if A and B are purely reactive with no resistance. In general elements A and B can carry their own resistances, so the simple subtraction fails and total R changes; the clean relation does not hold universally. The robust JEE-accepted answer for the general statement is 'None of these'.
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