In the given circuit, the charge on 4   μ F capacitor will be:

Question

In the given circuit, the charge on 4 &#8201; &#956; F capacitor will be:

Aicharya · Accepted Answer

The given circuit can be redrawn as:

Charge on $4 μ F$ capacitor:
$Q = (\frac{4 \times 6}{4 + 6}) \times 10$
$Q = 24 μC$

in series combination charge remains same.

so,

$Q_{4 μ F} = Q_{6 μ F} = 24 μ C$

In the given circuit, the charge on 4 &#8201; &#956; F capacitor will be: