Question

A single-loop circuit consists of a 7.1 Ω resistor, 11.8 H inductor, and a 3.2 μF capacitor. Initially the capacitor has a charge of 6.1 μC and the current is zero. Calculate the charge on the capacitor N complete cycles later for (a) N = 5, (b) N =10, and (c) N = 100.

Answer 1

Answer:

$q_{1} = 5.756\micro C$

$q_{2} = 5.43\micro C$

$q_{3} = 1.91\micro C$

Given:

Resistance, R = 7.1$\ohm$

Inductance, L = 11.8 H

Capacitance, C = 3.2$\micro F$

Initial charge in the capacitor, $q_{i} = 6.1\micro C$

Solution:

Charge is given by:

q = q'e$^{-t\frac{R}{2L}}cos(\omega t + \theta)$                          (1)

Now,

t = NT

T = $\frac{2\pi}{\omega}$

Thus

t = $\frac{2\pi N}{\omega}$

Substituting the above value of t in eqn (1):

q = q'e$^{- RN\frac{\frac{2\pi}{\omega}}{2L}}cos(\omega \frac{2\pi N}{\omega} + \theta)$

q = q'e$^{- RN\frac{\frac{2\pi}{\sqrt{\frac{L}{C}}}}{2L}}cos(2\pi N + \theta)$

q = q'e$^{- \pi RN\sqrt {\frac{C}{L}}} cos \theta$

where

$q'cos\theta = q_{i}$ = initial charge

q = q_{i}e$^{- \pi RN\sqrt {\frac{C}{L}}}$

Calculation of complete cycle for:

(a) N = 5

$q_{1} = 6.1\times 10^{-6}e^{- 5\time 7.1\pi \sqrt{\frac{3.2\times 10^{-6}}{11.8}}} = 5.756\micro C$

(b) N = 10

$q_{2} = 6.1\times 10^{-6}e^{- 10\time 7.1\pi \sqrt{\frac{3.2\times 10^{-6}}{11.8}}} = 5.43\micro C$

(c) N = 100

$q_{3} = 6.1\times 10^{-6}e^{- 100\time 7.1\pi \sqrt{\frac{3.2\times 10^{-6}}{11.8}}} = 1.91\micro C$