Question

A series circuit has a capacitor of 0.04×10−6 F and an inductor of 1 H. If the initial charge on the capacitor is 0.18×10−6C and there is no initial current, find the charge Q on the capacitor at any time t. Enter an exact answer. Do not use thousands separator in the answer field. Enclose arguments of functions in parentheses. For example, sin(2x).

Answer 1

Answer:

$Q(t) = 0.18\times 10^{-6} cos(5\times 10^3 t)$

Explanation:

We knwo that Kirchoff law

$L\frac{DI(t)}{dt} + \frac{1}{C}Q = 0$

where

$I(t) = \frac{dQ}{dt}$

hence

$LQ" + \frac{Q}{C} = 0$

C is given as 0.04\times 10^{6} F

L= 1 H , so we have

$Q" + 25\times 10^6Q = 0$

the characteristic equation of this differential equation is

$r^2 + 25\times 10^6 = 0$

$r = \pm 5\times 10^3 i$

Therefore differential equation is

$Q(t) = c_1 cos(5000t) + c_2sin(5000t)$

we know initial value if capacitor is given as $0.18\times 10^{-6} C$

Therefore

$0.18\times 10^{-6} = Q(0) = c_1 cos(0) + c_2sin(0)$

$c_1 = 0.18\times 10^{-6}$

if no inital current is present then we hvae I(0) = Q'(0) = 0

$Q'(T) = -5000 C_1 sin(5000t) + 5000 c_2cos(5000t)$

$0 = Q'(0) = - 5000c_1sin(0) + 5000c_2cos(0)$ therefre

$c_2 =0$

hence charge is

$Q(t) = 0.18\times 10^{-6} cos(5\times 10^3 t)$