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The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3 $Ω$ , 9 $Ω$ and 9 $Ω$ and a capacitor 5 $μ F$ .
How much is the current I in the circuit in steady state ?

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2.5 A

1.6 A

0.67 A

0.25 A

answer is D.

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Detailed Solution

In the steady state, capacitor will act as an open circuit. Therefore,

$I_{1} = I_{2} = I$

From the circuit diagram, we have
$8 + 3 I = 16 - 9 I$

$\Rightarrow I = 2 / 3 = 0.67 A$

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