PHYS208 Fundamentals of Physics II

Hint for Ch. 28, 43P

Refer to the LearningWare module on CD-Physics 2.0 for a systematic treatment of this circuit. Beware of the error in its second application of the loop rule!

An alternative to a fully general solution in terms of three currents i1, i2, and i3 relies on an underlying symmetry of the circuit. Note that the equivalent resistance of all three branches is 2.0 ohm. Thus since the emf in the center and right branch is identical, the same current must flow in each; by application of the junction rule, twice that current must flow in the left branch. An application of the loop rule to the left inner loop quickly yields the desired answers.

Hint for Ch. 28, 44P

Notice that by symmetry in the parallel configuration, the identical batteries must have identical currents. So the current in each battery is exactly one-half of the current in the load resistor.

In the series configuration, a quick application of the loop rule should convince you that an effective resistance of all three resistors in series and an effective emf of the sum of the two is acceptable.

Hint for Ch. 28, 48P

Refer to the multiloop example from class. It's in the same configuration with different circuit values.

Hint for Ch. 28, 57P

a) The 10 A through the internal resistance drops the terminal voltage of the battery from its emf value to 12 V.

b) The terminal voltage of the battery is now 9.6 V (8 A through 1.2 ohms of lights). Find current through battery.

Last updated Feb. 21, 1998.
Copyright George Watson, Univ. of Delaware, 1997.