Question

Two parallel plate capacitors of capacity 60 mF & 30 mF are charged to potentials of 40 V & 20 V, now these plates are joined by help of a battery of 80 V. Then final charge on positive plate of 60 mF

• A. 2400 mC
• B. 2800 mC
• C. 3200 mC
• D. 2000 mC

Answer 1

The correct option is B 2800 mC

Initial charge on plate of 60 mF$=60\times 40=2400$ mC.
Initial charge on plate of 30 mF$=30\times 20=600$ mC.
Let, x mC additionally be accumulated on positive plate of 60 mF. Then, by applying KVL
$80=\frac{x}{60}+40+\frac{x}{30}+20=\frac{3x}{60}+60$
$\Rightarrow x=\frac{\left(60\right)\left(20\right)}{3}=400mC$
Therefore total accumulated charge
$=2400+x=2800mC$