# Homework Help: Capacitor/inductor charging

1. Jun 2, 2009

### sirpsycho85

This isn't homework, but it feels like it...it's from the Dorf & Svoboda Introduction to Electric Circuits, 6th ed. on p 298, if you happen to have that.

I'm following the derivation of the formula for capacitor voltage for an RC circuit with a single resistor and single cap, or the current in the simple inductor circuit. Arriving at the general form of the differential equation, with time constant T:

dx(t)/dt + x(t)/T = K

rewritten:

dx/dt = (KT - x)/T

now, in the next step, the authors separate the variables and multiply each side by -1, yielding:

dx/(x - KT) = -dt/T

and go on to integrate both sides and arrive at:

ln(x - KT) = -t/T + D

D being the constant of integration. Raising e to both sides you get:

x(t) = KT + Ae-t/T

where A is eD, and you can go on using initial conditions to solve for the constant.

I can't get the same result when I don't multiply both sides by -1 when separating variables before solving the equation. My steps are as follows:

dx/dt = (KT - x)/T

dx/(KT - x) = dt/T having not multiplied by -1

ln(KT - x) = t/T + D

KT - x = Aet/T

x(t) = KT - Aet/T

This doesn't appear to be the same solution, as en is not equal to -e-n.

Oh, and hello everybody. I'm a junior engineer but work in a field that barely ever touches on a lot of what I learned in school...I've decided to start studying again to make sure I retain the fundamentals, especially if I do switch into a hardware design position.

Thank you.

2. Jun 3, 2009

### bartek2009

I think I spotted the mistake,

(d/dx)ln(x - KT) = 1/(KT - x)

but

(d/dx)ln(KT - x) = -1/(KT - x)

so

dx/(KT - x) ==> (-1)ln(KT - x)

hope that helps :)

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Last edited by a moderator: Aug 6, 2009
3. Jun 3, 2009

### sirpsycho85

You're right! I was thinking about the whole denominator when I should have been careful and integrated with respect to x.

Last edited: Jun 3, 2009