COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 18, Problem 79QAP
To determine

The power dissipated in each resistor.

Expert Solution & Answer
Check Mark

Answer to Problem 79QAP

  P1=45W

  P2=90W

  P3=28.125W

  P4=56.25W

Explanation of Solution

Given:

Resistors,

  R1=5ΩR2=10ΩR3=8ΩR4=16Ω

Voltage, V=45V

Formula:

According to Ohm's law the potential difference is given by,

  V=IRWhere,V=voltageI=currentR=resistance

Net resistance in series is given by,

  Rs=R1+R2Where,Rs=net resistanceR1 and R2=resistors connected in series

Net resistance in parallel is given by,

  1Rp=1R1+1R2Where,Rp=net resistanceR1 and R2=resistors connected in parallel

Charge in the conductor is given by,

  Q=ItWhere,Q=chargeI=currentt=time

Calculation:

Resistor R1 and R2 are connected in series. So, the net resistance in the series is

  Rs1=R1+R2Rs1=5+10Rs1=15Ω

Resistor R3 and R4 are connected in series. So, the net resistance in the series is

  Rs2=R3+R4Rs2=8+16Rs2=24Ω

Resistor Rs1andRs2 are connected in parallel. So, the net resistance in the parallel is

  1Rp=1R s 1+1R s 21Rp=115+124Rp=15×2415+24Rp=9.23Ω

Thus, net current in the circuit is,

  I=VRI=459.23I=4.875A

Now, current from Rs1 is,

  i1=VRs1=4515=3A

So, power dissipated in R1 is,

  P1=i12×R1P1=9×5P1=45W

Power dissipated in R2 is,

  P2=i12×R2P2=9×10P2=90W

Now, current from Rs2 is,

  i2=Ii2i2=4.8753i2=1.875A

Power dissipated in R3 is,

  P3=i22×R2P3=(1.875)2×8P3=28.125W

Power dissipated in R4 is,

  P4=i22×R2P4=(1.875)2×16P4=56.25W

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Chapter 18 Solutions

COLLEGE PHYSICS

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