Answer to Question #280613 in Differential Equations for Feline

Question

Given an RC series circuit that has an emf source of 110 volts, a resistance of 3 kilo-ohms, a capacitance of 5 microfarad and the initial charge of the capacitor is 50 micro coulomb. What is the charge in the circuit at the end of 0.05 seconds?

Answer 1

Differential equation of a LR circuit 

$$Rq'+\dfrac{q}{C}=U$$

Given $U=110V, R=3000\ Ohm, C=5\times10^{-6}\ F$

Substitute

$$3000q'+200000q=110$$ $$q'+\dfrac{200}{3}q=0.11$$

Integrating factor

$$\mu(t)=e^{\int(200/3)dt}=e^{(200/3)t}$$

$$e^{(200/3)t}(q'+\dfrac{200}{3}q)=0.11e^{(200/3)t}$$

$$d(e^{(200/3)t}q)=0.11e^{(200/3)t}dt$$

Integrate

$$\int d(e^{(200/3)t}q) =\int 0.11e^{(200/3)t}dt$$

$$e^{(200/3)t}q=0.00165e^{(200/3)t}+C$$

$$q(t)=1.65\times 10^{-3}+Ce^{-(200/3)t}$$

$q(0)=50\times 10^{-6} A$

$$0.05\times 10^{-3} =1.65\times 10^{-3}+C$$

$$C=-1.60\times 10^{-3}$$

$$q(t)=1.65\times 10^{-3}-1.60\times 10^{-3}e^{-(200/3)t}$$

$$q(0.05)=1.65\times 10^{-3}-1.60\times 10^{-3}e^{-(200/3)(0.05)}$$

$$q(0.05)=1.593\ mC$$