In a series AC circuit, the equations of applied voltage and the current flowing in the circuit are represented as
\( V = V_{0}\sin\omega t,\ i = i_{0}\sin(\omega t+\pi/4) \). The circuit may be
A. R-L circuit
B. R-C circuit
C. L-C circuit
D. L-C-R circuit
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → B and D only
In \( i = i_{0}\sin(\omega t+\pi/4) \) the current leads the voltage by \( \pi/4 \), which lies between 0 and \( \pi/2 \). The circuit must therefore contain resistance and be net capacitive.
\( \displaystyle \tan\phi = \frac{X_{C}-X_{L}}{R},\quad \phi = \frac{\pi}{4}\ \text{when}\ X_{C}-X_{L} = R \)
(A) R-L circuit — Incorrect. In an R-L circuit the current lags the voltage.
(B) R-C circuit — Correct. The current leads, with \( \tan\phi = X_{C}/R \); the lead is \( \pi/4 \) when \( X_{C} = R \).
(C) L-C circuit — Incorrect. With no resistance, the phase difference is exactly \( \pi/2 \) (leading or lagging), never \( \pi/4 \).
(D) L-C-R circuit — Correct. When \( X_{C} \gt X_{L} \) the current leads, and the lead is \( \pi/4 \) when \( X_{C}-X_{L} = R \).