Question

The circuit shown here has two batteries of $8.0\, V$ and $16.0\, V$ and three resistors $3 \Omega , 9 \Omega$ and $9 \Omega$ and a capacitor $5.0 \mu F$.

How much is the current $I$ in the circuit in steady state ?

• A. 1.6 A
• B. 0.67 A
• C. 2.5 A
• D. 0.25 A

Answer 1

Answer: (B)

In steady state capacitor is fully charged hence no current will flow through line $2$. By simplyfing the circuit
Hence resultant potential difference across resistances will be $8.0\, V.$
Thus current $I=\frac{V}{R}=\frac{8.0}{3+9}=\frac{8}{12}$
or, $I=\frac{2}{3}=0.67\,A$