Answer to Question #159156 in Electric Circuits for Ulrich

Question

A horizontal portion of a circuit contains 3 aligned points A, B and C. A 2.0uF and a 3.0uF capacitor are shunted between A and B, and there is a 5uF capacitor between B and C. Above this alignment is a series connection of a 100V battery and a switch S1 connected to A and C. Below the alignment is a 10 ohm resistor and a switch S2 connected to A and C. The switches are initially opened.
(i) Draw a diagram of the circuit
(ii) If S2 is left open and S1 closed, calculate the quantity of charge on each capacitor. How much energy does each store?

Answer 1

Explanations & Calculations

a) Circuit diagram

b)

• Since $S_2$ is kept open, no current flow through $\small 10\Omega$ resistor.
• What happens after $S_1$set-off is some amount of charge pass-through the capacitors & saturates (assume saturates) & after no charge pass-through.
• To calculate that amount, consideration of the equivalent capacitance would be easy.

$\qquad\qquad \begin{aligned} \small Equivalent &= \small \frac{5\mu F\times(2\mu F+3\mu F)}{5\mu F+(2\mu F+3\mu F)}\\ &= \small \frac{5}{2}\mu F\\ &= \small \bold{2.5\mu F} \end{aligned}$

• By $\small Q=CV$

$\qquad\qquad \begin{aligned} \small Q&= \small 2.5\mu F\times100V\\ &= \small 250\mu C \end{aligned}$

• Since the equivalent of the shunted two is also $\small 5\mu F$, that charge is qually distributes between the shunted two & $\small 5\mu F$.
• Therefore, the charge stored in $\small 5\mu F$ = $\small \bold{125\mu F}$

• The charge stored in $\small 2\mu F$ = $\small 125\mu F \times \large \frac{2}{(3+2)}=\small \bold{50\mu F}$
• In $\small 3\mu F$ = $\small 125-50=\bold{75\mu F}$

c)

• The energy on $\small 5\mu F$ = $\large \frac{Q^2}{2C}=\frac{(125\times10^{-6}C)^2}{2(5\times10^{-6}F)}=\small \bold{1.563\times10^{-3}J}$
• On $\small 2 \mu F$ =$\large \frac{(50\times10^{-6} C)^2}{2(2\times10^{-6}F)}=\small \bold{0.625\times10^{-3}J}$
• On $\small 3\mu F$ = $\large \frac{(75\times 10^{-6}C)^2}{2(3\times 10^{-6}F)}=\small \bold{0.938\times 10^{-3}J}$