Question 221 of 380

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


How much is the current $\mathrm{I}$ in the circuit in steady state?

A$1.6 \mathrm{~A}$B$0.67 \mathrm{~A}$C$2.5 \mathrm{~A}$D$0.25 \mathrm{~A}$

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Answer: (B) $0.67 \mathrm{~A}$

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 $8.0 \mathrm{~V}$.
Thus current $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{8.0}{3+9}=\frac{8}{12}$
or, $\mathrm{I}=\frac{2}{3}=0.67 \mathrm{~A}$