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JEE ADVANCED_2025

Paper-1 2025

Question

A circuit with an electrical load having impedance $Z$ is connected with an $A C$ source as shown in the diagram. The source voltage varies in time as $V(t)=300 \sin (400 t) V$, 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.

Select the correct option:

$\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$

$\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 3$

$\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1$

$\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 5$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$ \begin{aligned} & \phi=\tan ^{-1}\left(\frac{\omega L}{R}\right)=53^{\circ} \text { (lag) } \\ & (\mathrm{Q}) \rightarrow(5) \\ & (\mathrm{R}) Z=\sqrt{(30)^2+\left(\frac{1}{400 \times 50 \times 10^{-6}}-400 \times 25 \times 10^{-3}\right)^2} \\ & =\sqrt{(30)^2+(40)^2}=50 \Omega \\ & \therefore R \rightarrow(2) \end{aligned} $ (S) $ \begin{aligned} & Z=\sqrt{(60)^2+\left(\frac{1}{50 \times 10^{-6} \times 400}-125 \times 10^{-3} \times 400\right)} \\ & =\sqrt{(60)^2+(50-50)^2} \\ & =60 \Omega \\ & \therefore \quad i_0=\frac{300}{60}=5 \mathrm{~A}(S) \rightarrow 1 \end{aligned} $

Question Tags

JEE Advance

Physics

Easy

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