The circuit above is composed of a constant voltage source E, an inductor L, a resistor R and two switches S1 and S2. Prior to time t0 , both switches S1 and S2 are open.
a) Switch S1 is closed at time t0 = 0s. What is the current in the inductor L immediately after S1 was closed?
b) What is the current in the inductor L at a later time t1, after the voltage on resistor R1 has stabilized?
c) At time t2, after the current in R has stabilized, S2 is closed and S1 is opened at the same time. Write a differential equation and solve it to find the current in the inductor as a function of time passed from the moment t0, E, R, L and t2.
d) Sketch the shape of the current in the inductor L, from time t0 until the current stabilizes after the switch S2 is closed.
a) The current through the inductor L is zero immediately after S1 is closed. The current in an inductor is proportional with the integral of the voltage on the inductor.
b) If the current through the inductor has stabilized, di/dt = 0, and the voltage on the inductor is 0V:
uL = L(di/dt) = 0V.
According to the Kirchoff's Second law, E = 0 + RI and I = E/R.
c) Kirchoff's Second law:
Ldi/dt + Ri = 0
di/i = (-R/L)dt
ln(i) = (-R/L)t + C
At t = t2, the current is E/R, so ln(E/R) = (-R/L)t2 + C.
ln(i) = (-R/L)t + ln(E/R)+ + (R/L)t2
ln(Ri/E) = (-R/L)(t - t2)
Ri/E = e(-R/L)(t - t2)
i = (E/R)e-(t - t2)R/L for any t > t2
AP Physics C Practice