## PHYS208 Fundamentals of Physics II

### Quiz 9 -- LR Circuit

To prevent damage from arcing in an electric motor, a discharge resistor is sometimes placed in parallel with the armature. If the motor is suddenly unplugged while running, this resistor limits the voltage that appears across the armature coils.

Consider a 12 V dc motor that has an armature with a resistance of 7.5 ohm and an inductance of 450 mH. Assume the back emf in the armature coils is 10 V when the motor is running at normal speed.

a) Calculate the maximum resistance R that will limit the voltage across the armature to 80 V when the motor is unplugged.

b) How long does it take for the motor to de-energized? That is, after what time will the energy in the inductor be 99% dissipated?

Solution:

a) After the switch has been closed a long time, the current through the inductor is (12-10) V across 7.5 ohm. From Ohm's law the current through the inductor is 267 mA after a long time. You may also choose to apply Kirchhoff's loop rule to get this result.

When the switch is reopened, we require that the 267 mA through R generate only 80 V across it. It is in parallel with the armature! Again from Ohm's law, we require that R = 300 ohm.

b) The current in the inductor will be decaying exponentially with the inductive decay constant as the characteristic time. Note that you were asked about the energy, which is proportional to the square of the current. The time we seek is the value for which the following holds:

0.01 = exp(-2t/tauL)

The inductive time constant is the inductance, 450 mH, divided by the resistance in the left-most loop, 307.5 ohm; the time constant is 1.46 ms (150 ms if you forgot to include the 7.5 ohm). The solution of the equation above is t = (tauL/2) ln(100) = 3.36 ms. (If you considered decay of current instead of energy, your answer would be 6.72 ms.)
"http://www.physics.udel.edu/~watson/phys208/quiz9soln.html"
Last updated Nov. 21, 1997.
Copyright George Watson, Univ. of Delaware, 1997.