The Differential equation for RC-series circuit is
"R \\frac{dq}{dt} + \\frac{1}{C} q = E(t)" .
Now, "R = 500" ohm, C = "10^{-4\n}" F, E(t) = 100 V.
So, "500 \\frac{dq}{dt} + 10^4 q = 100 \\implies \\frac{dq}{dt} + 20 q = 0.2"
This is standard first-order differential equation.
So, Integrating factor = "e^{\\int20dt} = e^{20t}" .
Hence solution is "q e^{20t} = \\int(0.2)e^{20t} dt + c"
"\\implies qe^{20t} = (0.2) \\frac{e^{20t} }{20}+ c =0.01 e^{20t} +c"
Now, "q(0)=0 \\implies 0.01 + c = 0\\implies c = -0.01"
"\\implies qe^{20t} = 0.01(e^{20t}-1) \\\\ \\implies q = \\frac{0.01(e^{20t}-1)}{e^{20t}} = 0.01 (1-e^{-20t})"
And current "i(t) = \\frac{dq}{dt} = 0.2 e^{-20t}"
Comments
Dear Helen, thank you for correcting us.
Hi, Thank you for the solution :) I think you made a slight mistake though. Isn't i(t) supposed to be i(t)=0.2e^-20t . You forgot the -ve sign.
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