Question

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

• A. 17 A
• B. 7 A
• C. zero
• D. $\frac{10}{7}A$

Answer 1

Answer: (B)

7 A

Answer 2

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, $V=i\, R\, 10=\frac{i \times 10}{7}$ $\Rightarrow i=7\, A$