Determine Current in Circuit for Various Impedance Loads

To find the correct match between the electrical loads in List-I and the current functions $i(t)$ in List-II, we analyze the circuit for each given load connected to the AC source $V(t) = 300 \sin(400t)\text{ V}$.

From the source voltage equation:

• Peak voltage, $V_0 = 300\text{ V}$
• Angular frequency, $\omega = 400\text{ rad/s}$

The current as a function of time is given by: $i(t) = I_0 \sin(400t - \phi)$ where:

• $I_0 = \frac{V_0}{Z}$ is the peak current.
• $Z = \sqrt{R^2 + (X_L - X_C)^2}$ is the impedance of the circuit.
• $\phi = \arctan\left(\frac{X_L - X_C}{R}\right)$ is the phase angle by which the voltage leads the current.

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Analysis of Load (P):

• $R = 30\ \Omega$, $L = 0$, $C = 0$
• Impedance: $Z = R = 30\ \Omega$
• Peak current: $I_0 = \frac{300}{30} = 10\text{ A}$
• Phase angle: $\phi = 0$
• Current equation: $i(t) = 10 \sin(400t)\text{ A}$

Looking at List-II:

• Graph (3) has zero value at $400t = 0, \pi, 2\pi$, a peak of $+10\text{ A}$ at $400t = \pi/2$, and $-10\text{ A}$ at $400t = 3\pi/2$.

Thus, P $\rightarrow$ 3.

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Analysis of Load (Q):

• $R = 30\ \Omega$, $L = 100\text{ mH} = 0.1\text{ H}$
• Inductive reactance: $X_L = \omega L = 400 \times 0.1 = 40\ \Omega$
• Impedance: $Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = 50\ \Omega$
• Peak current: $I_0 = \frac{300}{50} = 6\text{ A}$
• Phase angle: $\tan\phi = \frac{X_L}{R} = \frac{40}{30} = \frac{4}{3} \implies \phi = \arctan\left(\frac{4}{3}\right) \approx 53^\circ$
• Current equation: $i(t) = 6 \sin(400t - 53^\circ)\text{ A}$

Key characteristics of this wave:

• At $400t = 0$: $i(0) = 6 \sin(-53^\circ) \approx -4.8\text{ A}$
• Zero crossing occurs when $400t = 53^\circ \approx 0.93\text{ rad}$ (between $0$ and $\pi/2$).
• Peak of $+6\text{ A}$ occurs at $400t = 90^\circ + 53^\circ = 143^\circ$ (between $\pi/2$ and $\pi$).

Looking at List-II:

• Graph (5) starts below zero at $400t = 0$, crosses zero before $\pi/2$, and reaches its peak between $\pi/2$ and $\pi$.

Thus, Q $\rightarrow$ 5.

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Analysis of Load (R):

• $C = 50\ \mu\text{F} = 50 \times 10^{-6}\text{ F}$, $R = 30\ \Omega$, $L = 25\text{ mH} = 25 \times 10^{-3}\text{ H}$
• Reactances: $X_C = \frac{1}{\omega C} = \frac{1}{400 \times 50 \times 10^{-6}} = 50\ \Omega$ $X_L = \omega L = 400 \times 25 \times 10^{-3} = 10\ \Omega$
• Net reactance: $X = X_L - X_C = 10 - 50 = -40\ \Omega$
• Impedance: $Z = \sqrt{R^2 + X^2} = \sqrt{30^2 + (-40)^2} = 50\ \Omega$
• Peak current: $I_0 = \frac{300}{50} = 6\text{ A}$
• Phase angle: $\tan\phi = \frac{-40}{30} = -\frac{4}{3} \implies \phi = -53^\circ$
• Current equation: $i(t) = 6 \sin(400t + 53^\circ)\text{ A}$

Key characteristics of this wave:

• At $400t = 0$: $i(0) = 6 \sin(53^\circ) \approx +4.8\text{ A}$
• Peak of $+6\text{ A}$ occurs at $400t = 90^\circ - 53^\circ = 37^\circ$ (before $\pi/2$).

Looking at List-II:

• Graph (2) starts near $+5\text{ A}$ at $400t = 0$ and reaches its peak before $\pi/2$.

Thus, R $\rightarrow$ 2.

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Analysis of Load (S):

• $C = 50\ \mu\text{F}$, $R = 60\ \Omega$, $L = 125\text{ mH} = 0.125\text{ H}$
• Reactances: $X_C = 50\ \Omega$ $X_L = \omega L = 400 \times 0.125 = 50\ \Omega$
• Net reactance: $X = X_L - X_C = 50 - 50 = 0\ \Omega\ \text{(Resonance)}$
• Impedance: $Z = R = 60\ \Omega$
• Peak current: $I_0 = \frac{300}{60} = 5\text{ A}$
• Phase angle: $\phi = 0$
• Current equation: $i(t) = 5 \sin(400t)\text{ A}$

Looking at List-II:

• Graph (1) is a sine wave with amplitude $5\text{ A}$ in phase with the source voltage.

Thus, S $\rightarrow$ 1.

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Final Matching:

$\text{P}\rightarrow 3, \quad \text{Q}\rightarrow 5, \quad \text{R}\rightarrow 2, \quad \text{S}\rightarrow 1$

This corresponds to option (A).