JEE Advance - Physics (2021 - Paper 1 Online - No. 8)
In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1 $$\mu$$C. Then S is switched to position Q. After a long time, the charge on the capacitor is q2 $$\mu$$C.
The magnitude of q2 is ________________.
The magnitude of q2 is ________________.
Answer
0.67
Explanation
Switch at position Q
$${V_A} - 1\,.\,{i_2} + 2 - 2{i_2} = {V_A}$$
$$3{i_2} = 2$$
$$ \Rightarrow {i_2} = {2 \over 3}A$$
Now, $${V_A} - \,{i_2} \times 1 = {V_B}$$
$${V_A} - {V_B} = {i_2} \times 1 = {2 \over 3}V$$
Potential difference across capacitor, $$\Delta V = {2 \over 3}V$$
$$\therefore$$ Charge on capacitor, $${q_2} = C\Delta V = 1 \times {2 \over 3} = 0.67\mu C$$
$${V_A} - 1\,.\,{i_2} + 2 - 2{i_2} = {V_A}$$
$$3{i_2} = 2$$
$$ \Rightarrow {i_2} = {2 \over 3}A$$
Now, $${V_A} - \,{i_2} \times 1 = {V_B}$$
$${V_A} - {V_B} = {i_2} \times 1 = {2 \over 3}V$$
Potential difference across capacitor, $$\Delta V = {2 \over 3}V$$
$$\therefore$$ Charge on capacitor, $${q_2} = C\Delta V = 1 \times {2 \over 3} = 0.67\mu C$$
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