Question 15

Given: $V(t) = 300 \sin(400t)$, so $V_{max} = 300 \text{ V}$ and $\omega = 400 \text{ rad/s}$.

(P) For a pure resistor $R = 30 \Omega$: $I_{max} = \frac{V_{max}}{R} = \frac{300}{30} = 10 \text{ A}$. The current is in phase with the voltage: $i(t) = 10 \sin(400t)$. This matches Graph (3).

(Q) For $R = 30 \Omega$ and $L = 100 \text{ mH}$: $X_L = \omega L = 400 \times 0.1 = 40 \Omega$. $Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = 50 \Omega$. $I_{max} = \frac{300}{50} = 6 \text{ A}$. Phase lag $\phi = \tan^{-1}(\frac{X_L}{R}) = \tan^{-1}(\frac{4}{3}) \approx 53^\circ$. The current lags the voltage. Graph (5) shows a lagging current with peak approximately $5-6$ A.

(R) For $C = 50 \mu\text{F}, R = 30 \Omega, L = 25 \text{ mH}$: $X_L = 400 \times 0.025 = 10 \Omega$. $X_C = \frac{1}{400 \times 50 \times 10^{-6}} = 50 \Omega$. $X_{net} = X_C - X_L = 40 \Omega$ (Capacitive). $Z = \sqrt{30^2 + 40^2} = 50 \Omega$. $I_{max} = \frac{300}{50} = 6 \text{ A}$. Phase lead $\phi = \tan^{-1}(\frac{40}{30}) = 53^\circ$. The current leads the voltage. Graph (2) shows a leading current with peak approximately $5-6$ A.

(S) For $C = 50 \mu\text{F}, R = 60 \Omega, L = 125 \text{ mH}$: $X_L = 400 \times 0.125 = 50 \Omega$. $X_C = 50 \Omega$. At resonance, $X_L = X_C$, so $Z = R = 60 \Omega$. $I_{max} = \frac{300}{60} = 5 \text{ A}$. The current is in phase with voltage. Graph (1) shows peak $5$ A and phase $0$.

Matching: P-3, Q-5, R-2, S-1. Correct Option: (A)