Question

5) Derive Equation 1 described in the handout of lab #7 by applying the loop rule to the de-energizing RL circuit, solving the differential equation to obtain the current function, and applying Ohm's law again.

Derive Equation 1 described in the handout of lab #7 by applying the loop rule to the de-energizing RL circuit, solving the differential equation to obtain the current function, and applying Ohm’s law again.

Answer 1

E = iRs + L di/dt (when the switch is closed)

di/dt + Rs/L i - E/L = 0

Solving this differntial equation, we get

I = E/Rs ( 1- e^(-Rst/L))
After a very long time, I = E/Rs

Now when the switch is re-opened

iRs + L di/dt = 0

di/dt + Rs/Li = 0

di/dt /i = -Rs/L
ln(i) = -Rs/L

Initial current = E/Rs

I = E/Rs e^(-Rst/L)
Vr = IRext
Vr = E*Rext/Rs e^(-Rst/L)