In the given circuit, a charge of +80 μ C is given to the upper plate of the 4 μ F capacitor. Then in the steady state, the charge on the upper plate of the 3 μ F capacitor is

Question

In the given circuit, a charge of +80 μ C is given to the upper plate of the 4 μ F capacitor. Then in the steady state, the charge on the upper plate of the 3 μ F capacitor is (A) +32 μ C  (B) +40 μ C  (C) +48 μ C  (D) +80 μ C

Tardigrade Admin · Accepted Answer

Between 3 μ F and 2 μ F (in parallel), total charge of 80 μ C will distribute in direct ratio of capacity. (q3/q2)= (3/2) q3= ( (3/3+2) )(80)=48 μ C

Q. In the given circuit, a charge of +80 $μ C$ is given to the upper plate of the 4 $μ F$ capacitor. Then in the steady state, the charge on the upper plate of the 3 $μ F$ capacitor is

$+ 32 μ C$

$+ 40 μ C$

$+ 48 μ C$

$+ 80 μ C$

Between $3 μ F$ and $2 μ F$ (in parallel), total charge of $80 μ C$ will distribute in direct ratio of capacity.
$\frac{q _{3}}{q _{2}} = \frac{3}{2}$
$q_{3} = (\frac{3}{3 + 2}) (80) = 48 μ C$

Q. In the given circuit, a charge of +80 μC is given to the upper plate of the 4 μF capacitor. Then in the steady state, the charge on the upper plate of the 3 μF capacitor is

+32μC

+40μC

+48μC

+80μC