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 $5.0\mu F.$ The current I in the circuit at steady state is :

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

Answer 1

The correct option is B 0.67 A

Answer is $B$
In the steady state, capacitor will act as an open circuit. Therefore,
$I_1=I_2=I$
From the circuit diagram, we have
$8+3I=16-9I$
Or, .
$I=2/3=0.67A$