Answer to Question #159159 in Electric Circuits for Bawe

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) Calculate the time constant and tge half-life of the circuit and state the significance of each value.

Answer 1

Explanations & Calculations

1)

2)

• The time constant is given by $\small \tau=RC \scriptsize(s)$: where R is the circuit resistance through which the capacitor either charged or discharged over time.
• Since there are 3 capacitors in this circuit, the equivalent capacitance should be calculated to represent this entire circuit, then the time constant becomes $\small \tau=RC_{equivalent}$

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

• For the discharging path of this circuit though $\small S_2\,\,and\,\,10\Omega$, resistance is $\small 10\Omega$. Then,

$\qquad\qquad \begin{aligned} \small \tau&= \small 10\Omega \times2.5\times10^{-6}F\\ &= \small 2.5\times10^{-5}s \end{aligned}$

• The time constant gives an idea of how much time it takes the circuit charge/discharge the capacitor to some amount. Readily it provides information about the time taken (at $\small t= RC$ ) for the capacitor to charge to 63.2% of the ultimate charge & time taken to discharge to 36.8% from the initially stored charge.
• This gives some idea of the response of the circuit over a change (transient response)

• Half-life is defined as the time taken for a quantity to become half of the initial. By the discharge equation $\small Q=Q_0 e^{-\frac{t}{RC}}$ of this circuit, the time that is taken for the charge to become half can be calculated.

$\qquad\qquad \begin{aligned} \small Q&\to\frac{Q_0}{2}\\ \small \frac{Q_0}{2}&= \small Q_0e^{-\frac{t}{RC}}\\ \small e^{\frac{t}{RC}}&= \small 2\\ \small t_{\frac{1}{2}}&= \small RC.ln2\\ &=\small 10\Omega\times2.5\times10^{-6}F\times ln2\\ &= \small \bold{1.733\times10^{-5}s} \end{aligned}$