College Physics

Example 21.2 Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a Parallel

Circuit

Let the voltage output of the battery and resistances in the parallel connection inFigure 21.4be the same as the previously considered series

connection:V= 12.0 V,R 1 = 1.00 Ω,R 2 = 6.00 Ω, andR 3 = 13.0 Ω. (a) What is the total resistance? (b) Find the total current.

(c) Calculate the currents in each resistor, and show these add to equal the total current output of the source. (d) Calculate the power dissipated

by each resistor. (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors.

Strategy and Solution for (a)

The total resistance for a parallel combination of resistors is found using the equation below. Entering known values gives

1 (21.21)

Rp

=^1

R 1

+^1

R 2

+^1

R 3

=^1

1.00 Ω

+^1

6.00 Ω

+^1

13.0 Ω

.

Thus,

1 (21.22)

Rp

=1.00

Ω

+^0.^1667

Ω

+^0.^07692

Ω

=^1.^2436

Ω

.

(Note that in these calculations, each intermediate answer is shown with an extra digit.)

We must invert this to find the total resistanceRp. This yields

R (21.23)

p=

1

1.2436

Ω = 0.8041 Ω.

The total resistance with the correct number of significant digits isRp= 0.804 Ω.

Discussion for (a)

Rpis, as predicted, less than the smallest individual resistance.

Strategy and Solution for (b)

The total current can be found from Ohm’s law, substitutingRpfor the total resistance. This gives

(21.24)

I= V

Rp

= 12.0 V

0.8041 Ω

= 14.92 A.

Discussion for (b)

CurrentIfor each device is much larger than for the same devices connected in series (see the previous example). A circuit with parallel

connections has a smaller total resistance than the resistors connected in series.

Strategy and Solution for (c)

The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage. Thus,

(21.25)

I 1 =V

R 1

= 12.0 V

1.00 Ω

= 12.0 A.

Similarly,

(21.26)

I 2 =V

R 2

=^12 .0 V

6.00 Ω

= 2.00 A

and

(21.27)

I 3 =V

R 3

= 12.0 V

13.0 Ω

= 0.92 A.

Discussion for (c)

The total current is the sum of the individual currents:

I 1 +I 2 +I 3 = 14.92 A. (21.28)

This is consistent with conservation of charge.

Strategy and Solution for (d)

The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three

are known. Let us useP=V

2

R

, since each resistor gets full voltage. Thus,

(21.29)

P 1 =V

2

R 1

=

(12.0 V)^2

1.00 Ω

= 144 W.

Similarly,

740 CHAPTER 21 | CIRCUITS, BIOELECTRICITY, AND DC INSTRUMENTS

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