In the circuit, shown in fig. $''K''$ is open. The charge on capacitor $C$ in steady state is $q_{1}.$ Now key is closed and at steady state, the charge on $C$ is $q_{2}.$ The ratio of charges $\left( \frac{q_{1}}{q_{2}} \right)\ $is [image:1_image73.emf]

Question

Dr. Abubacker · Accepted Answer

(a) When key is open, charge in steady state will be $q_{1} = CE$
When key is closed, potential difference across capacitor will be
$V = \frac{2R}{R + 2R}E = \frac{2}{3}E$
Charge in steady state will be $q_{2} = \frac{2}{3}CE \Rightarrow \frac{q_{1}}{q_{2}} = \frac{3}{2}$

Q. In the circuit, shown in fig. $''K''$ is open. The charge on capacitor $C$ in steady state is $q_{1}.$ Now key is closed and at steady state, the charge on $C$ is $q_{2}.$ The ratio of charges $\left( \frac{q_{1}}{q_{2}} \right)\ $is

Explanation

Reviewed by Dr. Abubacker

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