question_answer

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 of 5.0 \[\mu F.\]

How much is the current I in the circuit in steady state?                    [JEE ONLINE 12-04-2014]

A) 1.6 A

B) 0.67 A

C) 2.5 A

D) 0.25 A

Correct Answer: B

Solution :

[b]

In steady state capacitor is fully charged hence no current will flow through line 2.

By simplifying 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\[I=\frac{2}{3}=0.67A\]