JEE Advanced 2019 · Paper 1 · Physics Q7

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PhysicsMultiple correct+4 / -1Question 7 of 54

In the circuit shown, initially there is no charge on capacitors and keys $S_1$ and $S_2$ are open. The values of the capacitors are $C_1 = 10\ \mu$ F, $C_2 = 30\ \mu$ F and $C_3 = C_4 = 80\ \mu$ F.

Which of the statement(s) is/are correct?

Outer loop: from the top-left corner through key S₁ to point P, then C₄, a 5 V cell and a 30 Ω resistor to the top-right corner; down through C₃ to the bottom-right; along the bottom through 100 Ω to point Q and on to the bottom-left; up through 70 Ω and C₁ back to the top-left. A middle branch from P to Q contains key S₂, C₂, a 30 Ω resistor and a 10 V cell.

At time

t = 0

, the key

S_1

is closed, the instantaneous current in the closed circuit will be 25 mA.

If key

S_1

is kept closed for long time such that capacitors are fully charged, the voltage across the capacitor

C_1

will be 4 V.

The key

S_1

is kept closed for long time such that capacitors are fully charged. Now key

S_2

is closed, at this time, the instantaneous current across 30 Ω resistor (between points P and Q) will be 0.2 A (round off to 1st decimal place).

If key

S_1

is kept closed for long time such that capacitors are fully charged, the voltage difference between points P and Q will be 10 V.

Answer and worked solution

Official answer: A, B

With $S_1$ closed and $S_2$ open, the outer loop contains $C_1$ , $70\ \Omega$ , $100\ \Omega$ , $C_3$ , $30\ \Omega$ , the 5 V cell and $C_4$ .

(A) At $t=0$ the uncharged capacitors act as shorts:
$I=\frac5{70+100+30}=25\text{ mA}.$
True.

(B) In the steady state $C_1$ , $C_4$ and $C_3$ are in series:
$\frac1C=\frac1{10}+\frac1{80}+\frac1{80}\implies C=8\ \mu\text{F},\qquad Q=40\ \mu\text{C}.$
So $V_{C_1}=40/10=4$ V. True.

(D) No current flows, so the bottom wire, including Q, is at one potential. P is connected to Q only through $C_1$ (and the idle resistors), so $|V_{PQ}|=V_{C_1}=4$ V, not 10 V. False.

(C) At the instant $S_2$ closes, the rest of the network acts between P and Q like a 4 V source with internal resistance $70\parallel130\approx45.5\ \Omega$ . The 10 V cell in the middle branch opposes it, so
$I=\frac{10-4}{30+45.5}\approx0.08\text{ A},$
not 0.2 A. False.

(A), (B)

Topics:RC circuits Series and parallel combinations Kirchhoff's laws

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