The circuit shown here has two batteries of $8.0 V$ and $16.0 V$ and three resistors $3Ω, 9Ω$ and $9Ω$ and a capacitor of $5.0 μ F$ .
How much is the current $I$ in the circuit in steady state?

Question

The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3Ω , 9Ω and 9Ω and a capacitor of 5.0 μ F . How much is the current I in the circuit in steady state?

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

Physics - Capacitor Question with Solution | TestHub

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