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.

Question

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. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Initial charge on C1 is Q1 = C1 V= 110 μ C Let x charge flow through wires. (Q1 - x/C1) = (x/Ceq) where Ceq = (C2 × C3/C2 + C3) Solve to get x = 60 μ C.

Initial charge on C1​ is Q1​=C1​V=110μC Let x charge flow through wires. C1​Q1​−x​=Ceq​x​ where Ceq​=C2​+C3​C2​×C3​​ Solve to get x=60μC.

Initial charge on $C_{1}$ is $Q_{1} = C_{1} V = 110 μ C$
Let $x$ charge flow through wires.
$\frac{Q _{1} - x}{C _{1}} = \frac{x}{C _{e q}}$
where $C_{e q} = \frac{C _{2} \times C _{3}}{C _{2} + C _{3}}$
Solve to get $x = 60 μ C$ .