JEE MAIN - Physics (2025 - 29th January Evening Shift - No. 18)
A capacitor, $C_1 = 6 \mu F$ is charged to a potential difference of $V_0 = 5V$ using a 5V battery. The battery is removed and another capacitor, $C_2 = 12 \mu F$ is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges ($q_1$ and $q_2$) on the capacitors $C_1$ and $C_2$ when equilibrium condition is reached.
_29th_January_Evening_Shift_en_18_1.png)
Explanation
Given, $${c_1} = 6\mu F,{c_2} = 12\mu F,v = 5v$$
We know, for a capacitor $$Q = CV$$
_29th_January_Evening_Shift_en_18_2.png)
Initial charge on $${c_1}\,.\,Q = {c_1}v$$
$$ \Rightarrow Q = 6 \times 5$$
$$ \Rightarrow Q = 30\mu C$$ .... (1)
when switch 'S' is closed, let q charge flows from C$_1$ to C$_2$
So final charges are $Q-q$ and q on C$_1$ and C$_2$ respectively.
Now, when equilibrium condition is reached potential on C$_1$ and C$_2$ becomes equal.
Hence, $${{Q - q} \over {{c_1}}} = {q \over {{c_2}}}$$ (As $$v = {Q \over c}$$)
$$ \Rightarrow {Q \over {{c_1}}} = q\left( {{{{c_1} + {c_2}} \over {{c_1}{c_2}}}} \right)$$
$$ \Rightarrow q = \left( {{{{c_2}} \over {{c_1} + {c_2}}}} \right)Q$$
$$ \Rightarrow q = \left( {{{12} \over {6 + 12}}} \right) \times 30\mu C$$ (From (1))
$$ \Rightarrow q = \left( {{{12} \over {18}}} \right) \times 30 = 20\mu c$$
and $$Q - q = 30 - 20 = 10\mu c$$
Hence, $$q = 10\mu c$$ and $${q_2} = 20\mu c$$
Comments (0)
