question_answer

In the circuit here, the steady state voltage across capacitor C is a fraction of the battery e.m.f. The fraction is decided by                                                                             [AMU (Engg.) 2000]

A)            \[{{R}_{\text{1}}}\] only

B)            \[{{R}_{\text{1}}}\] and \[{{R}_{\text{2}}}\] only

C)            \[{{R}_{1}}\] and \[{{R}_{3}}\] only

D)            \[{{R}_{1}}\],\[{{R}_{2}}\] and \[{{R}_{3}}\]

Correct Answer: B

Solution :

                   In steady state potential difference across capacitor \[{{V}_{\text{2}}}\]= potential difference across resistance \[{{R}_{2}}=\left( \frac{{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}} \right)\,V\] Hence ,\[{{V}_{\text{2}}}\] depends upon \[{{R}_{\text{2}}}\] and \[{{R}_{\text{1}}}\]