Question

Condensers with capacities $C,2C,3C$ and $4C$ are chaged to the voltage $V,2V,3V$ and $4V$ respectively (figure). The circuit is closed. Find the voltage on all the condensers in equilibrium.

Answer 1

FOR IMAGE 02:

let after closing the switch current $q$ starts flowing through the circuit.

By applying Kirchhoff law in closed loop.

$\Rightarrow \frac{q_1-q}{C}+\frac{q_2-q}{2C}+\frac{q_3-q}{3C}+\frac{q_4-q}{4C}=0$

$\Rightarrow \frac{CV-q}{C}+\frac{4CV-q}{2C}+\frac{9CV-q}{3C}+\frac{16CV-q}{4C}=0$

$\Rightarrow [V+2V+3V+4V]-[\frac{q}{C}+\frac{q}{2C}+\frac{q}{3C}+\frac{q}{4C}]=0$

$\left[10V\right]-\left(\frac{24+12+8+6}{24C}\right)q=0$

$\Rightarrow q=\frac{24}{5}CV$

Voltage across capacitors:

${\left|△V\right|}_C=\left|\frac{q_1-q}{C}\right|=\left|\frac{CV-24/5CV}{C}\right|=\frac{19}{5}V$

${\left|△V\right|}_{2}C=\left|\frac{q_2-q}{2C}\right|=\left|\frac{4CV-24/5CV}{2C}\right|=\frac{2}{5}V$

${\left|△V\right|}_{3}C=\left|\frac{q_3-q}{3C}\right|=\left|\frac{9CV-24/5CV}{3C}\right|=\frac{7}{5}V$

${\left|△V\right|}_{4}C=\left|\frac{q_4-q}{4C}\right|=\left|\frac{16CV-24/5CV}{4C}\right|=\frac{14}{5}V$