question_answer

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

A)  \[17\,A\]

B)  \[7\,A\]

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, \[V=iR\] \[10=\frac{i\times 10}{7}\] \[\Rightarrow \] \[i=7A\]