COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 18, Problem 83QAP
To determine

(a)

The equivalent resistance of given figure.

Expert Solution
Check Mark

Answer to Problem 83QAP

The equivalent resistance of the circuit is 16Ω.

Explanation of Solution

Given:

The given circuit is shown below.

  COLLEGE PHYSICS, Chapter 18, Problem 83QAP , additional homework tip  1

Figure 1
Resistors are as,
  R1=20Ω

  R2=12Ω

  R3=8Ω

Formula used:

For series combination, equivalent resistance can be written as,
  Req=R1+R2+R3+......

For parallel combination of resistors, equivalent resistance can be written as,
  1Req=1R1+1R2+1R3+......

Calculation:

R1 and R2 are in parallel combination, so equivalent resistance of R1 and R2 will be,
  1R 12=1R1+1R21R 12=120Ω+112Ω=3+560Ω=860Ω=215ΩR12=152Ω=7.5Ω

Now, R12 and R3 are in series, so, equivalent resistance can be written as,
  Req=R12+R3Req=7.5Ω+8Ω=15.5Ω16Ω

Conclusion:

Thus, equivalent resistance of the given circuit is 16Ω.

To determine

(b)

The current through each resistor

Expert Solution
Check Mark

Answer to Problem 83QAP

Current through R1=20Ω is 0.12A.

Current through R2=12Ω is 0.20A.

Current through R3=8Ω is 0.32A.

Explanation of Solution

Given:

Potential difference between a and b = Vab=5V

  R1=20Ω

  R2=12Ω

  R3=8Ω

Formula used:

By ohm's law,
  V=iR

V = potential difference between two points
  i = electric current
R = resistance
Calculation:

Total current = itotal=VabReq=5V15.5Ω=0.32A

From the figure 1, total current through the circuit = current through R3=8Ω

So, itotal=i3=0.32A

  COLLEGE PHYSICS, Chapter 18, Problem 83QAP , additional homework tip  2

Again, from the figure
  Vab=Vac+VcbVab=Vac+i3R3Vac=Vabi3R3Vac=5V(0.32A×8Ω)Vac=5V2.56VVac=2.44V

Now, by ohm's law,
  i1=V abR1=2.44V20Ω=0.122A=0.12Ai2=V abR2=2.44V12Ω=0.20333A=0.20A

Conclusion:

Thus, the current through R1=20Ω, R2=12Ω and R3=8Ω are 0.12A,0.20A and 0.32A respectively.

To determine

(c)

The power dissipated in each resistor.

Expert Solution
Check Mark

Answer to Problem 83QAP

Power dissipated in R1=20Ω is 0.29W

Power dissipated in R2=12Ω is 0.49W

Power dissipated in R3=8Ω is 0.84W

Explanation of Solution

Given:

From part b,
Current through R1=20Ω is 0.12A.

Current through R2=12Ω is 0.20A.

Current through R3=8Ω is 0.32A.

Formula used:

Power dissipated across resistance,
  P=i2Ri= electric currentR = electric resistance

Calculation:

Power dissipated across R1=20Ω,
  P1=i12R1=(0.12A)2×20Ω=0.29W

Power dissipated across R2=12Ω,
  P2=i22R2=(0.2033A)2×12Ω=0.49W

Power dissipated across R3=8Ω,
  P1=i12R1=(0.324A)2×8Ω=0.84W

Conclusion:

Power dissipated in R1=20Ω is 0.29W

Power dissipated in R2=12Ω is 0.49W

Power dissipated in R3=8Ω is 0.84W

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Chapter 18 Solutions

COLLEGE PHYSICS

Ch. 18 - Prob. 11QAPCh. 18 - Prob. 12QAPCh. 18 - Prob. 13QAPCh. 18 - Prob. 14QAPCh. 18 - Prob. 15QAPCh. 18 - Prob. 16QAPCh. 18 - Prob. 17QAPCh. 18 - Prob. 18QAPCh. 18 - Prob. 19QAPCh. 18 - Prob. 20QAPCh. 18 - Prob. 21QAPCh. 18 - Prob. 22QAPCh. 18 - Prob. 23QAPCh. 18 - Prob. 24QAPCh. 18 - Prob. 25QAPCh. 18 - Prob. 26QAPCh. 18 - Prob. 27QAPCh. 18 - Prob. 28QAPCh. 18 - Prob. 29QAPCh. 18 - Prob. 30QAPCh. 18 - Prob. 31QAPCh. 18 - Prob. 32QAPCh. 18 - Prob. 33QAPCh. 18 - Prob. 34QAPCh. 18 - Prob. 35QAPCh. 18 - Prob. 36QAPCh. 18 - Prob. 37QAPCh. 18 - Prob. 38QAPCh. 18 - Prob. 39QAPCh. 18 - Prob. 40QAPCh. 18 - Prob. 41QAPCh. 18 - Prob. 42QAPCh. 18 - Prob. 43QAPCh. 18 - Prob. 44QAPCh. 18 - Prob. 45QAPCh. 18 - Prob. 46QAPCh. 18 - Prob. 47QAPCh. 18 - Prob. 48QAPCh. 18 - Prob. 49QAPCh. 18 - Prob. 50QAPCh. 18 - Prob. 51QAPCh. 18 - Prob. 52QAPCh. 18 - Prob. 53QAPCh. 18 - Prob. 54QAPCh. 18 - Prob. 55QAPCh. 18 - Prob. 56QAPCh. 18 - Prob. 57QAPCh. 18 - Prob. 58QAPCh. 18 - Prob. 59QAPCh. 18 - Prob. 60QAPCh. 18 - Prob. 61QAPCh. 18 - Prob. 62QAPCh. 18 - Prob. 63QAPCh. 18 - Prob. 64QAPCh. 18 - Prob. 65QAPCh. 18 - Prob. 66QAPCh. 18 - Prob. 67QAPCh. 18 - Prob. 68QAPCh. 18 - Prob. 69QAPCh. 18 - Prob. 70QAPCh. 18 - Prob. 71QAPCh. 18 - Prob. 72QAPCh. 18 - Prob. 73QAPCh. 18 - Prob. 74QAPCh. 18 - Prob. 75QAPCh. 18 - Prob. 76QAPCh. 18 - Prob. 77QAPCh. 18 - Prob. 78QAPCh. 18 - Prob. 79QAPCh. 18 - Prob. 80QAPCh. 18 - Prob. 81QAPCh. 18 - Prob. 82QAPCh. 18 - Prob. 83QAPCh. 18 - Prob. 84QAPCh. 18 - Prob. 85QAPCh. 18 - Prob. 86QAPCh. 18 - Prob. 87QAPCh. 18 - Prob. 88QAPCh. 18 - Prob. 89QAPCh. 18 - Prob. 90QAPCh. 18 - Prob. 91QAPCh. 18 - Prob. 92QAP
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