General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 17, Problem 27E

(a)

To determine

The current and power dissipated in the circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 27E

The current in the circuit is 3000A_ and power dissipated is 3.6×104W_.

Explanation of Solution

Given that the voltage is 12V, the internal resistance of the battery is 0.004ohm, and the resistance of the conductor is negligibly small.

Write the expression for the current according to Ohm’s law.

  I=VR        (I)

Here, V is the voltage, I is the current, and R is the resistance.

Write the expression for power dissipated.

  P=IV        (II)

Here, P is the power.

Conclusion:

Substitute 12V for V, and 0.004ohm for R in equation (I) to find I.

  I=12V0.004ohm=3000A

Substitute 3000A for I, and 12V for V in equation (II) to find P.

  P=(3000A)(12V)=3.6×104W

Therefore, the current in the circuit is 3000A_ and power dissipated is 3.6×104W_.

(b)

To determine

The resistance that required to draw 80A current.

(b)

Expert Solution
Check Mark

Answer to Problem 27E

The resistance that required to draw 80A current is 0.146ohm_.

Explanation of Solution

Given that the current drawn is 80A.

Equation (I) can be solved for R to find the net resistance of the circuit required to draw the specified amount of current.

  R=VI        (III)

Since the battery has internal resistance, the required resistance can be computed as,

  Rrequired=RRinternal        (IV)

Here, Rrequired is the resistance required, Rinternal is the internal resistance of the battery.

Conclusion:

Substitute 12V for V, and 80A for I in equation (III) to find R.

  R=12V80A=0.15ohm

Substitute 0.15ohm for R, and 0.004ohm for Rinternal in equation (IV) to find Rrequired.

  Rrequired=0.15ohm0.004ohm=0.146ohm

Therefore, the resistance that required to draw 80A current is 0.146ohm_.

(c)

To determine

The power dissipated in the motor.

(c)

Expert Solution
Check Mark

Answer to Problem 27E

The power dissipated in the motor is 934W_.

Explanation of Solution

Given that the current drawn by the motor is 80A, and the resistance is 0.146ohm.

Write the expression for the power supplied.

  P=I2R        (V)

Here, P is the power.

Conclusion:

Substitute 80A for I, and 0.146ohm for R in equation (V) to find P.

  P=(80A)2(0.146ohm)=934W

Therefore, the power dissipated in the motor is 934W_.

(d)

To determine

The power dissipated in the battery.

(d)

Expert Solution
Check Mark

Answer to Problem 27E

The power dissipated in the battery is 25.6W_.

Explanation of Solution

Given that the internal resistance of the battery is 0.004ohm.

Equation (V) can be used to find the power dissipated in the battery, by replacing R by the internal resistance Rinternal.

  P=I2Rinternal        (VI)

Conclusion:

Substitute 80A for I, and 0.004ohm for Rinternal in equation (VI) to find P.

  P=(80A)2(0.004ohm)=25.6W

Therefore, the power dissipated in the battery is 25.6W_.

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Chapter 17 Solutions

General Physics, 2nd Edition

Ch. 17 - Prob. 11RQCh. 17 - Prob. 12RQCh. 17 - Prob. 1ECh. 17 - Prob. 2ECh. 17 - Prob. 3ECh. 17 - Prob. 4ECh. 17 - Prob. 5ECh. 17 - Prob. 6ECh. 17 - Prob. 7ECh. 17 - Prob. 8ECh. 17 - Prob. 9ECh. 17 - Prob. 10ECh. 17 - Prob. 11ECh. 17 - Prob. 12ECh. 17 - Prob. 13ECh. 17 - Prob. 14ECh. 17 - Prob. 15ECh. 17 - Prob. 16ECh. 17 - Prob. 17ECh. 17 - Prob. 18ECh. 17 - Prob. 19ECh. 17 - Prob. 20ECh. 17 - Prob. 21ECh. 17 - Prob. 22ECh. 17 - Prob. 23ECh. 17 - Prob. 24ECh. 17 - Prob. 25ECh. 17 - Prob. 26ECh. 17 - Prob. 27ECh. 17 - Prob. 28ECh. 17 - Prob. 29ECh. 17 - Prob. 30ECh. 17 - Prob. 31ECh. 17 - Prob. 32ECh. 17 - Prob. 33ECh. 17 - Prob. 34ECh. 17 - Prob. 35ECh. 17 - Prob. 36ECh. 17 - Prob. 37ECh. 17 - Prob. 38ECh. 17 - Prob. 39ECh. 17 - Prob. 40ECh. 17 - Prob. 41ECh. 17 - Prob. 42ECh. 17 - Prob. 43ECh. 17 - Prob. 44ECh. 17 - Prob. 45ECh. 17 - Prob. 46ECh. 17 - Prob. 47ECh. 17 - Prob. 48ECh. 17 - Prob. 49ECh. 17 - Prob. 50ECh. 17 - Prob. 51ECh. 17 - Prob. 52ECh. 17 - Prob. 53ECh. 17 - Prob. 54ECh. 17 - Prob. 55ECh. 17 - Prob. 56ECh. 17 - Prob. 57ECh. 17 - Prob. 58ECh. 17 - Prob. 59ECh. 17 - Prob. 60ECh. 17 - Prob. 61ECh. 17 - Prob. 62ECh. 17 - Prob. 63ECh. 17 - Prob. 64ECh. 17 - Prob. 65ECh. 17 - Prob. 66ECh. 17 - Prob. 67ECh. 17 - Prob. 68ECh. 17 - Prob. 69ECh. 17 - Prob. 70ECh. 17 - Prob. 71ECh. 17 - Prob. 72ECh. 17 - Prob. 73ECh. 17 - Prob. 74ECh. 17 - Prob. 75ECh. 17 - Prob. 76ECh. 17 - Prob. 77ECh. 17 - Prob. 78ECh. 17 - Prob. 79ECh. 17 - Prob. 80ECh. 17 - Prob. 81ECh. 17 - Prob. 82ECh. 17 - Prob. 83ECh. 17 - Prob. 84ECh. 17 - Prob. 85ECh. 17 - Prob. 86ECh. 17 - Prob. 87ECh. 17 - Prob. 88ECh. 17 - Prob. 89ECh. 17 - Prob. 90ECh. 17 - Prob. 91ECh. 17 - Prob. 92E
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