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?

Question

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?

Aicharya · Accepted Answer

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, $\mathrm{I}=\frac{2}{3}=0.67 \mathrm{~A}$

The circuit shown here has two batteries of $8.0 \mathrm{~V}$ and $16.0 \mathrm{~V}$ and three resistors $3…

Solution