A circuit with an electrical load having impedance Z is connected with an AC source as shown in the diagram. The source voltage varies in time as V(t) = 300 sin(400t) V, where t is time in s. List-I shows various options for the load. The possible currents i(t) in the circuit as a function of time are given in List-II.

Choose the option that describes the correct match between the entries in List-I to those in List-II.

The given AC source voltage is:
$V\left(t\right)=300sin\left(400t\right)V$
From this expression, we can identify the following system parameters:
- Peak source voltage: $V_0=300V$
- Angular frequency: $\omega =400rad/s$

Let us analyze each load configuration in List-I one by one:

1. Analysis of Load (P):
In the load diagram (P), a capacitor of capacitance $C=50\mu F$ and an inductor of inductance $L=25mH$ are bypassed (short-circuited) by a connecting wire, leaving only the resistor $R=30\Omega$ in the active circuit path.
Thus, the load is purely resistive:
$Z=R=30\Omega$
For a purely resistive circuit, the current and voltage are in phase ($\phi =0$).
The peak current is:
$I_0=\frac{V_0}{Z}=\frac{300}{30}=10A$
The instantaneous current is:
$i\left(t\right)=10sin\left(400t\right)A$
This corresponds to graph (3) in List-II (in-phase waveform with peak value of 10 A).
Hence, P → 3.

2. Analysis of Load (Q):
In configuration (Q), a resistor $R=30\Omega$ and an inductor $L=100mH$ are connected in series.
The inductive reactance is:
$X_L=\omega L=400\times 100\times {10}^{-3}=40\Omega$
The total impedance of this RL series circuit is:
$Z=\sqrt{R^2+X_L^2}=\sqrt{{30}^2+{40}^2}=50\Omega$
The peak current is:
$I_0=\frac{V_0}{Z}=\frac{300}{50}=6A$
In an RL circuit, the current lags the voltage by a phase angle $\phi$:
$tan\phi =\frac{X_L}{R}=\frac{40}{30}=\frac{4}{3}\Rightarrow \phi \approx 53°$
Thus, the instantaneous current is:
$i\left(t\right)=6sin(400t-53°)A$
At $t=0$, $i\left(0\right)=6sin(-53°)\approx -4.8A$.
This corresponds to graph (5) in List-II, which has a peak of 6 A and is negative at $t=0$.
Hence, Q → 5.

3. Analysis of Load (R):
In configuration (R), a capacitor $C=50\mu F$, resistor $R=30\Omega$, and inductor $L=25mH$ are connected in series.
The reactances are:
$X_C=\frac{1}{\omega C}=\frac{1}{400\times 50\times {10}^{-6}}=50\Omega$
$X_L=\omega L=400\times 25\times {10}^{-3}=10\Omega$
The net reactance is:
$X=X_C-X_L=50-10=40\Omega$ (capacitive)
The impedance of this RLC circuit is:
$Z=\sqrt{R^2+{(X_C-X_L)}^2}=\sqrt{{30}^2+{40}^2}=50\Omega$
The peak current is:
$I_0=\frac{V_0}{Z}=\frac{300}{50}=6A$
Since $X_C>X_L$, the current leads the voltage by a phase angle $\phi$:
$tan\phi =\frac{X_C-X_L}{R}=\frac{40}{30}=\frac{4}{3}\Rightarrow \phi \approx 53°$
Thus, the instantaneous current is:
$i\left(t\right)=6sin(400t+53°)A$
At $t=0$, $i\left(0\right)=6sin(53°)\approx 4.8A$.
This matches graph (2) in List-II (peak of 6 A and positive at $t=0$).
Hence, R → 2.

4. Analysis of Load (S):
In configuration (S), a capacitor $C=50\mu F$, resistor $R=60\Omega$, and inductor $L=125mH$ are connected in series.
The reactances are:
$X_C=\frac{1}{\omega C}=\frac{1}{400\times 50\times {10}^{-6}}=50\Omega$
$X_L=\omega L=400\times 125\times {10}^{-3}=50\Omega$
Since $X_L=X_C=50\Omega$, the circuit is in resonance.
The impedance is purely resistive:
$Z=R=60\Omega$
The phase difference is $\phi =0$ because the net reactance is zero.
The peak current is:
$I_0=\frac{V_0}{Z}=\frac{300}{60}=5A$
The instantaneous current is:
$i\left(t\right)=5sin\left(400t\right)A$
This corresponds to graph (1) in List-II (in-phase waveform with peak value of 5 A).
Hence, S → 1.

Combining these matches gives:
P → 3, Q → 5, R → 2, S → 1