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

Let 'q' be the final charge on 3 μ F capacitor then (80- q /2)=( q /3) ⇒ q =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

$+ 32 μ C$

$+ 40 μ C$

$+ 48 μ C$

$+ 80 μ C$

Let ' $q$ ' be the final charge on $3 μ F$ capacitor then
$\frac{80 - q}{2} = \frac{q}{3} \Rightarrow q = 48 μ C$

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

+32μC

+40μC

+48μC

+80μC