Q47 / 48 · JEE Advanced 2025 (Paper 1) · Physics

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 $V(t)=300 \sin (400 t) \mathrm{V}$, 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.
Figure
Choose the option that describes the correct match between the entries in List-I to those in List-II.
List-I List-II
$(P)$ Figure
 $(1)$ Figure
$(Q)$ Figure
 $(2)$ Figure
$(R)$ Figure
 $(3)$ Figure
$(S)$ Figure
 $(4)$ Figure
$(5)$ Figure

  • Correct answer
    A
    $\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$
  • B
    $\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 3$
  • C
    $\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$
  • D
    $\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 5$

Solution

For P $i=\frac{V}{R}=10 \sin 400 t \Rightarrow(3)$ For Q $\begin{aligned} & X_L=\omega L=400 \times 100 \times 10^{-3}=40 \Omega \\ & \therefore Z=50 \Omega \\ & \left.\therefore i=\frac{300}{50} \sin \left(400 t-53^{\circ}\right) \text { [current will lag by } \tan ^{-1} \frac{X_L}{R}\right] \Rightarrow(5)\end{aligned}$ For R $\begin{aligned} & \mathrm{X}_{\mathrm{C}}=\frac{10^6}{400 \times 50} \Omega=50 \Omega \text { and } \mathrm{X}_{\mathrm{L}}=400 \times 25 \times 10^{-3}=10 \Omega \\ & \therefore \mathrm{Z}=50 \Omega \\ & \therefore \mathrm{i}=\frac{300}{50} \sin \left(400 \mathrm{t}+53^{\circ}\right) \quad\left[\text { Current will lead by } \tan ^{-1} \frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\right] \Rightarrow(2)\end{aligned}$ For $S$ $\therefore \mathrm{i}=\frac{300}{60} \sin (400 \mathrm{t}) \quad \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \Rightarrow$ Resonance $\Rightarrow(1)$

Hence, the correct answer is option: 'A'.

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