Answer to Question #272277 in Differential Equations for Mizzy

Question

Given an RC series circuit that has an emf source of 50 volts, a resistance of 20k ohms, a capacitance of 6 microfarad and the initial charge of the capacitor is 1 microcoulomb. What is the charge in the capacitor at the end of 0.01 second? What is the current in the circuit at the end of 0.05 seconds?

Answer 1

$$Rq'+\dfrac{1}{C}q=V$$

$$2\cdot10^{4}q'+\dfrac{1}{6\cdot10^{-6}}q=50$$

$$q'+\dfrac{25}{3}q=0.0025$$

$$q'=-\dfrac{25}{3}(q-0.0006)$$

$$\dfrac{dq}{q-0.0003}=-\dfrac{25}{3}dt$$

Integrate

$$q=0.0003+c_1e^{-25t/3}$$

Given $q(0)=10^{-6}$

$$0.000001=0.0003+c_1$$

$$c_1=-0.000299$$

$$q(t)=0.0003-0.000299e^{-25t/3}$$

$$q(0.01)=0.0003-0.000299e^{-25(0.01)/3}$$

$$=24.91\cdot10^{-6} (C)$$

$24.91\ \mu C$

$$i(t)=q'(t)=-0.000299(-\dfrac{25}{3})e^{-25t/3}$$

$$=\dfrac{0.007475}{3}e^{-25t/3}$$

$$i(0.05)=\dfrac{0.007475}{3}e^{-25(0.05)/3}=0.0016426(A)$$

$$=1.64(\ mA)$$

$1.64\ mA$