JEE Advanced · Physics · 22. AC Circuits
In the circuit shown, initially there is no charge on capacitors and keys and are open. The values of the capacitors are and

Which of the statement(s) is/are correct?
- A If key is kept closed for long time such that capacitors are fully charged, the voltage difference between points and will be
- B The keys is kept closed for long time such that capacitors are fully charged. Now key is closed, at this time, the instantaneous current across resistor (between points and ) will be (round off to decimal place).
- C At time the key is closed, the instantaneous current in the closed circuit will be
- D If key is kept closed for long time such that capacitors are fully charged, the voltage across the capacitors will be
Answer & Solution
Correct Answer
(D) If key is kept closed for long time such that capacitors are fully charged, the voltage across the capacitors will be
Step-by-step Solution
Detailed explanation
For option , refer circuit diagram below
Just after closing of switch charge on all the capacitors
Replace all capacitors with wire, then circuit will get reduced as shown below.
is correct
Now is kept closed for long time circuit is in steady state
Apply KVL in loop, start from battery
Potential difference across
is correct.
For option Now just after closing of charge on each capacitor remain same,
Refer circuit below -
KVL in left loop, starting from battery of volts,
...(i)
Now, apply KVL in right loop starting from 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,
and is incorrect.
Just after closing of switch charge on all the capacitors
Replace all capacitors with wire, then circuit will get reduced as shown below.
is correct
Now is kept closed for long time circuit is in steady state
Apply KVL in loop, start from battery
Potential difference across
is correct.
For option Now just after closing of charge on each capacitor remain same,
Refer circuit below -
KVL in left loop, starting from battery of volts,
...(i)
Now, apply KVL in right loop starting from 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,
and is incorrect.
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