Question

Two capacitors of $2\mu F$ and $3\mu F$ are charged to $150V$ and $120V$ respectively. The plates of capacitor are connected as shown in the figure. An uncharged capacitor of capacity $1.5\mu F$ falls to the free end of the wire. Then :

• A. charge of $1.5\mu F$ capacitor is $180\mu C$
• B. charge on $2\mu F$ capacitor is $120\mu C$
• C. positive charge flows through A from right to left
• D. positive charge flows through A from left to right

Answer 1

Answer: (A), (B), (D)

charge of $1.5\mu F$ capacitor is $180\mu C$
charge on $2\mu F$ capacitor is $120\mu C$
positive charge flows through A from left to right

Initially charge on $2\mu F=2\times 150=300\mu C$ and charge on $3\mu F=3\times 120=360\mu C$

When the uncharged capacitor of capacity $1.5\mu F$ falls to the free end of the wire, it will make a closed loop circuit. Let $q\mu C$ charge flows in the closed loop in clockwise direction. Then final charges on different capacitors are as shown in figure.

Applying Kirchhoff's law for the loop,

$\frac{300-q}{2}-\frac{q}{1.5}+\frac{360-q}{3}=0$

multiplying by 30 both sides, $4500-15q-20q+3600-10q=0$

or $8100=45q\Rightarrow q=180\mu C$

Thus, positive charge flow through A from left to right.

charge on $2\mu F$ becomes $=300-180=120\mu C$