31.71. Model: The battery and the connecting wires are ideal.
Visualize:
The figure shows how to simplify the circuit in Figure P31.71 using the laws of series and parallel resistances.
Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit
to find the current and potential difference of each resistor.
Solve: (a) From the last circuit in the figure and from Kirchhoff’s law, I = 100 V 10 Ω = 10 A . Thus, the current
through the battery is 10 A. Now as we rebuild the circuit, we note that series resistors must have the same current I
and that parallel resistors must have the same potential difference.
In Step 1 of the above diagram, we return the 10 Ω resistor to the 4 Ω, 4 Ω, and 2 Ω resistors in series. These
resistors must have the same 10 A current as the 10 Ω resistance. That is, the current through the 2 Ω and the 4 Ω
resistors is 10 A. The potential differences are
∆V2 = (10 A)(2 Ω) = 20 V
∆V4 (left) = (10 A)(4 Ω) = 40 V
∆V4 (left) = (10 A)(4 Ω) = 40 V
In Step 2, we return the left 4 Ω resistor to the 20 Ω and 5 Ω resistors in parallel. The two resistors must have the
same potential difference ∆V = 40 V. From Ohm’s law,
I5 =
40 V
= 8A
5Ω
I20 =
40 V
=2A
20 Ω
The currents through the various resistors are I2 = I4 = 10 A, I5 = 8 A, and I20 = 2 A.
2
(b) The power dissipated by the 20 Ω resistor is I20
(20 Ω) = (2 A)2 (20 Ω) = 80 W .
(c) Starting with zero potential at the grounded point, we travel along the outside path to point a and add/subtract
the potential differences on the way:
0 V + (20 Ω)I20 + (2 Ω)I2 = (20 Ω)(2 A) + (2 Ω)(10 A) = 60 V = Va