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 $1.5\mu F$ is connected to the free end of the wires as shown. Then:

• A. Charge on $1.5\mu F$ 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)

The correct options are
A Charge on $1.5\mu F$ capacitor is $180\mu C$
B Charge on $2\mu F$ capacitor is $120\mu C$
D Positive charge flows through $A$ from left to right

The initial charges were
$q_{2\mu C}=2\left(150\right)=300\mu C$
$q_{3\mu C}=3\left(150\right)=360\mu C$
$q_{1.5\mu C}=0$

Let the charges on the capacitors after the connection be, as shown.


Now plates (1) and (3) are isolated
$\Rightarrow q+q_1=\mathrm{300...}\left(1\right)$
Plates (4) and (5) are isolated
$\Rightarrow -q_1+q_2=-300+360$
$-q_1+q_2=\mathrm{60...}\left(2\right)$
Plates (2) and (6) are isolated
$-q-q_2=-360$
$q+q_2=\mathrm{360...}\left(3\right)$

$KVL$ gives
$-\frac{q}{1.5}+\frac{q_2}{3}+\frac{q_1}{2}=0$
$-4q+2q_2+3q_1=0$
$-4q+2(360-q)+3(300-q)=0\left(\text{from}\right(1\left)\text{and}\right(3\left)\right)$
$-4q+2(360-q)+3(300-q)=0$
$\Rightarrow q=180\mu C$
$\Rightarrow q_1=120\mu C,q_2=180\mu C$

Flow of charge from $2\mu F$ capacitor is $q_{2\mu C}-q_1=180\mu C$