Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Chapter 28, Problem 18SP

(a)

To determine

The equivalent resistance of 8 Ω and 4 Ω resistances in series.

(a)

Expert Solution
Check Mark

Answer to Problem 18SP

Solution:

12 Ω

Explanation of Solution

Given data:

The resistance of the first resistor is 8 Ω.

The resistance of the second resistor is Ω.

Formula used:

The expression for equivalent resistance in series is written as:

Req=R1+R2

Here, R1 and R2 are the resistances of the resistor in series and Req is the equivalent resistance.

Explanation:

Recall the expression for equivalent resistance in series.

Req=R1+R2

Substitute 8 Ω for R1 and Ω for R2

 Req=8 Ω+4 Ω=12 Ω

Conclusion:

The equivalent resistance is 12 Ω.

(b)

To determine

The equivalent resistance of 8 Ω and 4 Ω resistances in parallel.

(b)

Expert Solution
Check Mark

Answer to Problem 18SP

Solution:

The equivalent resistance is 2.7 Ω.

Explanation of Solution

Given data:

The resistance of the first resistor is 8 Ω.

The resistance of the second resistor is Ω.

Formula used:

The expression for equivalent resistance in parallel is written as:

1Req=1R1+1R21Req=R1+R2R1R2Req=R1R2R1+R2

Here, R1 and R2 are the resistances of the resistor in parallel and Req is the equivalent resistance.

Explanation:

Recall the expression for equivalent resistance in parallel.

Req=R1R2R1+R2

Substitute 8 Ω for R1 and Ω for R2

Req=(8 Ω)(4 Ω)8 Ω+4 Ω =3212=2.7 Ω

Conclusion:

The equivalent resistance is 2.7 Ω.

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