Consider the circuit sketched above. The battery has emf of 27 volts. The induct

Question

Consider the circuit sketched above. The battery has emf of 27 volts. The inductance is L = 0.40 H and the resistances are R_1 = 12 Ohm and R_2 = 9.0 Ohm. Use the figure to answer the following. Initially the switch S is open and no currents flow. Then the switch is closed. What is the current in the resistor R_1 just after the switch is closed? After leaving the switch closed for a long time, it is opened again. Just after it is opened, what is the current in R_1? After leaving the switch closed for a long time, it is opened again. How long does it take for the current to reach 1/3 of the maximum current.

Explanation / Answer

a) An inductor resists a change in current. After the switch is closed, the inductor will momentarily look like an open circuit.
So

i2 = 0 i1 = E/R1 = 27/12 = 2.25 A

The current through the switch is i1 = 2.25 A

Since i2 = 0 V(R2) = 0V

V(L) = E = 27 V

Using V(L) = L*di/dt we get di/dt = V(L)/L = 27V/0.4 = 67.5 A/s

b) Open switch. An inductor does not like any change in current through it. Steady state, it has been conducting 2.25 amps. That will not change immediately when the switch opens.

Current through R1 will remain the same at 2.25 amps. Current through R2 will also be 2.25 amps, but will have reversed direction (being supplied by the inductor now instead of the battery).

c) Tc = L/R

R = R1 + R2 = 21 ohms

Tc = 0.4/21 = 0.019 s
I(t) = I(m)*(1-e^(-t/Tc))

I(m) = 2.25 A

I(t) = I(m) / 3

1-I(t)/I(m) = e^(-t/Tc)
-ln(1-I(t)/I(m)) = t/Tc
t = -Tc*ln(1-I(t)/I(m))

t = -Tc*ln(1/9) = 0.0417 s

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