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JEE Advanced · Physics · 22. AC Circuits

In the circuit shown, initially there is no charge on capacitors and keys S1 and S2 are open. The values of the capacitors are C1=10 μF, C2=30 μF and C3= C4=80 μF.

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

  1. A If key S1 is kept closed for long time such that capacitors are fully charged, the voltage difference between points P and Q will be 10 V.
  2. B The keys S1 is kept closed for long time such that capacitors are fully charged. Now key S2 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).
  3. C At time t=0, the key S1 is closed, the instantaneous current in the closed circuit will be 25 mA.
  4. D If key S1 is kept closed for long time such that capacitors are fully charged, the voltage across the capacitors C1 will be 4V.
Verified Solution

Answer & Solution

Correct Answer

(D) If key S1 is kept closed for long time such that capacitors are fully charged, the voltage across the capacitors C1 will be 4V.

Step-by-step Solution

Detailed explanation

For option (C) , refer circuit diagram below

Just after closing of switch charge on all the capacitors C1, C3 and C4=0
Replace all capacitors with wire, then circuit will get reduced as shown below.

i=570+100+30=5200=25 mA(C) is correct
Now S1 is kept closed for long time circuit is in steady state

Apply KVL in loop, start from battery
+5-280-210-280=0
10.280=5
2=40 μC
Potential difference across C1
ΔVC1=210=4010=4 volt
(D) is correct.
For option (B) Now just after closing of S2 charge on each capacitor remain same,
Refer circuit below -

KVL in left loop, starting from battery of 10 volts,
10-30x-4010-70y=0
30x+70y=6 ...(i)
Now, apply KVL in right loop starting from 80 μF capacitor
\(\Rightarrow-\frac{40}{80}+5+(x-y) 30-\frac{40}{80}+(x-y) \times 100\) \(-10+x \times 30=0 \)
\( \Rightarrow 160 x-130 y-6=0 \ldots(\text{ii})\)
Now, solving equation (i) and (ii) we get,
y=961510
and x=0.05 amp. (B) is incorrect.
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