Alternating Current: Exam-Pattern Drill (7)

Question

An AC circuit has $R = 30\,\Omega$, $X_L = 50\,\Omega$, and $X_C = 10\,\Omega$ in series. The applied voltage is $V_{rms} = 200\text{ V}$. Find the current and power factor.

Solution — Step by Step

Final answer: $I_{rms} = 4\text{ A}$, $\cos\phi = 0.6$

Why This Works

In a series RLC circuit, only resistance dissipates power. Inductors and capacitors store and return energy each cycle — they shift the phase but consume nothing on average. The power factor $\cos\phi = R/Z$ measures how much of the total impedance is “useful” (resistive).

A 3-4-5 triangle hides in this problem: $R = 30$, net reactance = 40, $Z = 50$. JEE loves Pythagorean triples.

Alternative Method

Compute $\tan\phi = (X_L - X_C)/R = 40/30 = 4/3$, then $\cos\phi = 3/5 = 0.6$. Same answer via trigonometry.

Common Mistake