College Physics

motor and a passenger compartment light. We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects,

as the next example indicates.

Example 21.3 Calculating Resistance,IRDrop, Current, and Power Dissipation: Combining Series and Parallel

Circuits

Figure 21.6shows the resistors from the previous two examples wired in a different way—a combination of series and parallel. We can consider

R 1 to be the resistance of wires leading toR 2 andR 3. (a) Find the total resistance. (b) What is theIRdrop inR 1? (c) Find the currentI 2

throughR 2. (d) What power is dissipated byR 2?

Figure 21.6These three resistors are connected to a voltage source so thatR 2 andR 3 are in parallel with one another and that combination is in series withR 1.

Strategy and Solution for (a)

To find the total resistance, we note thatR 2 andR 3 are in parallel and their combinationRpis in series withR 1. Thus the total (equivalent)

resistance of this combination is

Rtot=R 1 +Rp. (21.34)

First, we findRpusing the equation for resistors in parallel and entering known values:

1 (21.35)

Rp

=^1

R 2

+^1

R 3

=^1

6.00 Ω

+^1

13.0 Ω

=0.2436

Ω

.

Inverting gives

(21.36)

Rp=^1

0.2436

Ω = 4.11 Ω.

So the total resistance is

Rtot=R 1 +Rp= 1.00 Ω + 4.11 Ω = 5.11 Ω. (21.37)

Discussion for (a)

The total resistance of this combination is intermediate between the pure series and pure parallel values (20.0 Ωand0.804 Ω, respectively)

found for the same resistors in the two previous examples.

Strategy and Solution for (b)

To find theIRdrop inR 1 , we note that the full currentIflows throughR 1. Thus itsIRdrop is

V 1 =IR 1. (21.38)

We must findIbefore we can calculateV 1. The total currentIis found using Ohm’s law for the circuit. That is,

(21.39)

I= V

Rtot

= 12.0 V

5.11 Ω

= 2.35 A.

Entering this into the expression above, we get

V 1 =IR 1 = (2.35 A)(1.00 Ω ) = 2.35 V. (21.40)

Discussion for (b)

The voltage applied toR 2 andR 3 is less than the total voltage by an amountV 1. When wire resistance is large, it can significantly affect the

operation of the devices represented byR 2 andR 3.

Strategy and Solution for (c)

To find the current throughR 2 , we must first find the voltage applied to it. We call this voltageVp, because it is applied to a parallel

combination of resistors. The voltage applied to bothR 2 andR 3 is reduced by the amountV 1 , and so it is

742 CHAPTER 21 | CIRCUITS, BIOELECTRICITY, AND DC INSTRUMENTS

This content is available for free at http://cnx.org/content/col11406/1.7