FlipIt for College Physics (Algebra Version - Six Months Access)
FlipIt for College Physics (Algebra Version - Six Months Access)
17th Edition
ISBN: 9781319032432
Author: Todd Ruskell
Publisher: W.H. Freeman & Co
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Chapter 18, Problem 69QAP
To determine

The value of R1, R2, and R3 in circuit.

Expert Solution & Answer
Check Mark

Answer to Problem 69QAP

R1, R2, and R3 are 27-ohm, 108 ohm and 108 ohm.

Explanation of Solution

Given:

Voltage present in circuit, Vb= 9 V

Power dissipated in each resistor, P1= P2 = P3 = 1.50 W

Formula used: Power dissipates across resistor

  P = V2R=I2R.

Where, P is power dissipated

V = Voltage across resistor

I = Current flowing through resistor

R = Resistance..

Equivalent resistance, when resistance connects in series

  Req=Ra+Rb.

Equivalent resistance, when resistance connects in parallel

  1Req=1Ra+1Rb.

Equivalent power, when resistance connects in series

  1Peq=1Pa+1Pb.

Equivalent power, when resistance connects in parallel

  Peq=Pa+Pb.

Figure:

FlipIt for College Physics (Algebra Version - Six Months Access), Chapter 18, Problem 69QAP , additional homework tip  1.

FlipIt for College Physics (Algebra Version - Six Months Access), Chapter 18, Problem 69QAP , additional homework tip  2.

FlipIt for College Physics (Algebra Version - Six Months Access), Chapter 18, Problem 69QAP , additional homework tip  3.

Calculation:

In figure (1)

Power in each resistance is equal..

  P2=P3(Vbe)2R2=(Vcd)2R3bothareconnectinparallel,Vbe=VcdR2=R3......(a).

Equivalent Power of R2and R3

  (P23)eq=P2+P3(P23)eq=1.5W+1.5W(P23)eq=3W......(b).

Equivalent resistance of R2and R3

  1(R23)eq=1R2+1R3R2=R3(R23)eq=R22......(c).

In figure (2)

Current flowing through resistanceR1and(Req)23is I1 Power of both resistanceR1and(Req)23.

  I12R1=P1I12R1=1.5W.......(d)and,I12(Req)23=(P23)eqI12(R22)=3W.......(e).

On comparing equation (d) and (e)

  R2=4R1......(f).

So, equivalent resistance of circuit

  Req=R1+(R23)eqReq=R1+R22R2=4R1Req=R1+4R12Req=3R1......(g).

Equivalent power of circuit

  1Peq=1P1+1(P23)eq1Peq=11.5W+13WPeq=1W.......(h).

In figure (3)

  Peq=(Veq)2ReqReq=3R11W=(9V)23R1R1=27Ω

From equation (f),

  R2=4R1=108Ω.

From equation (a),

  R3=R2=108Ω.

Conclusion:

Thus, the value ofR1=27ΩandR2=R3=108Ω.

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Chapter 18 Solutions

FlipIt for College Physics (Algebra Version - Six Months Access)

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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