question_answer

At \[t=0\], switch S is closed. The charge on the capacitor is varying with time as \[Q={{Q}_{0}}(1-{{e}^{-\alpha t}})\] Find the value of \[{{Q}_{0}}\].

A) \[\frac{CV{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\]          

B) \[\frac{CV{{R}_{1}}}{{{R}_{1}}+{{R}_{2}}}\]

C) \[\frac{CV{{R}_{1}}{{R}_{2}}}{({{R}_{1}}-{{R}_{2}})({{R}_{1}}+{{R}_{2}})}\]

D)    None

Correct Answer: A

Solution :

                \[{{Q}_{0}}=\] Steady state charge = C (potential difference across capacitor or resistor\[{{R}_{2}}\] in steady state) \[=C\left( \frac{V}{{{R}_{1}}+{{R}_{2}}} \right){{R}_{2}}=\frac{CV{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\]