Answer to Question #280613 in Differential Equations for Feline

Question #280613

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?

1
Expert's answer
2021-12-20T15:35:53-0500

Differential equation of a LR circuit 


Rq+qC=URq'+\dfrac{q}{C}=U

Given U=110V,R=3000 Ohm,C=5×106 FU=110V, R=3000\ Ohm, C=5\times10^{-6}\ F

Substitute


3000q+200000q=1103000q'+200000q=110q+2003q=0.11q'+\dfrac{200}{3}q=0.11

Integrating factor


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

e(200/3)t(q+2003q)=0.11e(200/3)te^{(200/3)t}(q'+\dfrac{200}{3}q)=0.11e^{(200/3)t}

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

Integrate


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

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

q(t)=1.65×103+Ce(200/3)tq(t)=1.65\times 10^{-3}+Ce^{-(200/3)t}

q(0)=50×106Aq(0)=50\times 10^{-6} A

0.05×103=1.65×103+C0.05\times 10^{-3} =1.65\times 10^{-3}+C

C=1.60×103C=-1.60\times 10^{-3}

q(t)=1.65×1031.60×103e(200/3)tq(t)=1.65\times 10^{-3}-1.60\times 10^{-3}e^{-(200/3)t}

q(0.05)=1.65×1031.60×103e(200/3)(0.05)q(0.05)=1.65\times 10^{-3}-1.60\times 10^{-3}e^{-(200/3)(0.05)}

q(0.05)=1.593 mCq(0.05)=1.593\ mC


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