JEE Advanced · Physics · 22. AC Circuits
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.
Choose the option that describes the correct match between the entries in List-I to those in List-II.
| List-I | List-II | ||
|---|---|---|---|
| \((P)\) |  | \((1)\) |  |
| \((Q)\) |  | \((2)\) |  |
| \((R)\) |  | \((3)\) |  |
| \((S)\) |  | \((4)\) |  |
| \((5)\) |  |
- 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 \(\mathrm{P} \rightarrow 1, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 5\)
Answer & Solution
Correct Answer
(A) \(\mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 1\)
Step-by-step Solution
Detailed explanation
For P
\(i=\frac{V}{R}=10 \sin 400 t \Rightarrow(3)\)
For Q
\(X_L=\omega L=400 \times 100 \times 10^{-3}=40 \Omega \)
\( \therefore Z=50 \Omega \)
\(\therefore i=\frac{300}{50} \sin \left(400 t-53^{\circ}\right)\) \([\)current will lag by \(\tan ^{-1} \frac{X_L}{R}] \Rightarrow(5)\)
For R
\(\mathrm{X}_{\mathrm{C}}=\frac{10^6}{400 \times 50} \Omega=50 \Omega \text { and } \mathrm{X}_{\mathrm{L}}=400 \times 25 \times\) \(10^{-3}=10 \Omega \)
\( \therefore \mathrm{Z}=50 \Omega \)
\( \therefore \mathrm{i}=\frac{300}{50} \sin \left(400 \mathrm{t}+53^{\circ}\right) \quad\) \([\)Current will lead by \(\tan ^{-1} \frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}] \Rightarrow(2)\)
For \(S\)
\(\therefore \mathrm{i}=\frac{300}{60} \sin (400 \mathrm{t}) \quad \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \Rightarrow\) Resonance \(\Rightarrow(1)\)
\(i=\frac{V}{R}=10 \sin 400 t \Rightarrow(3)\)
For Q
\(X_L=\omega L=400 \times 100 \times 10^{-3}=40 \Omega \)
\( \therefore Z=50 \Omega \)
\(\therefore i=\frac{300}{50} \sin \left(400 t-53^{\circ}\right)\) \([\)current will lag by \(\tan ^{-1} \frac{X_L}{R}] \Rightarrow(5)\)
For R
\(\mathrm{X}_{\mathrm{C}}=\frac{10^6}{400 \times 50} \Omega=50 \Omega \text { and } \mathrm{X}_{\mathrm{L}}=400 \times 25 \times\) \(10^{-3}=10 \Omega \)
\( \therefore \mathrm{Z}=50 \Omega \)
\( \therefore \mathrm{i}=\frac{300}{50} \sin \left(400 \mathrm{t}+53^{\circ}\right) \quad\) \([\)Current will lead by \(\tan ^{-1} \frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}] \Rightarrow(2)\)
For \(S\)
\(\therefore \mathrm{i}=\frac{300}{60} \sin (400 \mathrm{t}) \quad \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \Rightarrow\) Resonance \(\Rightarrow(1)\)
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