Question

A 14.1 μF capacitor is charged to a potential of 45.0 V and then discharged through a 75.0 Ω resistor. (a) How long after discharge begins does it take for the capacitor to lose 90.0% of the following?

(i) its initial charge? _______s

(ii) its initial energy? ______s

(b) What is the current through the resistor at both times in part (a)?

(i) at t_charge ? _________ A

(ii) at t_energy ? ________A

Answer 1

Data Given

C = 41.1 uF, V = 45 V, R = 75 Ohm

When capacitor is fully charged charge on the capacitor

$Q_{0} = C V = 14.1\times 10^{-6}\times 45 = 6.345\times 10^{-4}C$

Time constant for the circuit

$\tau = RC = 75\times 14.1\times 10^{-6}s = 1.0575 \times 10^{-3}s$

We know that Now during discharge

$Q = Q_{0}e^{\frac{-t}{\tau }}$

Part (a) Charge left on the capaciter Q = 100%-90% = 10 % of Q₀

10 100

   Taking log on bot sides

e 1.0575×10-35 575x10's 100

$\frac{-t}{1.0575\times 10^{-3}s} = \ln \left ( \frac{10}{100} \right ) = \ln 10 - \ln 100$

$\frac{-t}{1.0575\times 10^{-3}s} = -2.3026$

$t = 2.3026\times 1.0575\times 10^{-3}s = 2.435\times 10^{-3}s$

Part a(ii) When it lost 90% of its initial energy

Initial Energy of the capacitor

$E_{i}= \frac{1}{2}CV^{2} = \frac{1}{2}\times 14.1\times 10^{-6}\times 45^{2}= 14276.25 \times 10^{-6} J$

Energy left on capacitor when it lost 90% of E_i

$E_{f}= 10\% Of E_{i}= \frac{10}{100}\times 14276.25\times 10^{-6} =1427.625\times 10^{-6} J$

$\frac{Q^{2}}{2C}=1427.625\times 10^{-6} J$

$Q^{2}=1427.625\times 10^{-6} J\times 2C = 1427.625\times 2\times 14.1\times 10^{-12}C^{2}$

$Q= 200.65\times 10^{-6}C$

Where $Q_{0}= 14.1\times 45 = 634.5 \times 10^{-6}C$

Again using

$\frac{200.65\times 10^{-6}}{634.5\times 10^{-6}} = e^{\frac{-t}{1.0575 \times 10^{-3}s }}$ taking log on bot sides

$\frac{-t}{1.0575 \times 10^{-3}s } = \ln \left ( \frac{200.65}{634.5} \right ) = -1.15$

t-1.15 × 1.0575 × 10-3s 1.216125 × 10-3 S

Part B) Current Flow in first case

li = VIT = 45 × є1.0435x10-3, = 0.0599.4 = 0.06 × e 1.0575×10-') 5 ー 75

In second Caes

12 = Veユ= 45 × e-1057525 ol T, = 0.01899-1-0.02A ー×e 75 1.0575×10 ,