question_answer

The steady state current through the battery in the circuit given below is

A)  17A                                      

B)  7A

C)  zero                                     

D) \[\frac{10}{7}A\]

Correct Answer: B

Solution :

                 In steady state, the branch containing capacitor will become ineffective (ie, we can ignore\[1\,\Omega ,\]resistance) The equivalent circuit will now be \[\therefore \]  \[{{R}_{eq}}=\frac{{{R}_{1}}\times {{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{5\times 2}{5+2}=\frac{10}{7}\] Now,        \[V=iR\]                 \[10=\frac{i\times 10}{7}\] \[\Rightarrow \]               \[i=7\text{ }A\]