1. Tardigrade
  2. Question
  3. Physics
  4. A capacitor of capacitance C1 = 1 μ F charged up to a voltage V = 110 V is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing capacitances C2 = 2 μ F and C3 = 3 μ F. Then, the amount of charge (in μ C) that will flow through the connecting wires is.

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Q. A capacitor of capacitance $C_1 = 1 \,\mu F$ charged up to a voltage $V = 110 \,V$ is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing capacitances $C_2 = 2\, \mu F$ and $C_3 = 3\, \mu F$. Then, the amount of charge (in $\mu C)$ that will flow through the connecting wires is_______.

Electrostatic Potential and Capacitance

Solution:

Initial charge on $C_1$ is $Q_1 = C_1 V= 110\, \mu C$
Let $x$ charge flow through wires.
$\frac{Q_1 - x}{C_1} = \frac{x}{C_{eq}}$
where $C_{eq} = \frac{C_2 \times C_3}{C_2 + C_3}$
Solve to get $ x = 60\,\mu C$.