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 32E

(a)

To determine

The EMF ε in the given circuit.

(a)

Expert Solution
Check Mark

Answer to Problem 32E

The EMF ε in the given circuit is 6V_.

Explanation of Solution

The circuit is shown in Figure 17.25. Let us represent the 1ohm resistance as R1, 3ohm resistance as R2, 2V battery as V2, and the current as I.

The current is flowing in the anti-clockwise direction in the loop. Apply Kirchhoff’s voltage law to the given circuit.

  V2IR2IR1ε=0        (I)

Solve equation (I) for ε.

  ε=V2IR1IR2        (II)

Conclusion:

Substitute 2V for V2, 2A for I, 1ohm for R1, and 3ohm for R2 in equation (II) to find ε.

  ε=2V(2A)(1ohm)(2A)(3ohm)=6V|ε|=6V

Therefore, the EMF ε in the given circuit is 6V_.

(b)

To determine

The power supplied to each of the two resistors.

(b)

Expert Solution
Check Mark

Answer to Problem 32E

The power supplied to the 1ohm resistor is 4W_, and that to the 3ohm resistor is 12W_

Explanation of Solution

Given that the current through each of the resistors is 2A.

Write the expression for the power supplied to a resistor.

  P=I2R        (III)

Here, P is the power, I is the current, and R is the resistance.

The power supplied to each the resistors can be computed from equation (III) as,

  P1=I2R1P2=I2R2}        (IV)

Conclusion:

Substitute 2A for I, 1ohm for R1, and 3ohm for R2 in equation (IV) to find the power supplied resistors.

  P1=(2A)2(1ohm)=4W

  P2=(2A)2(3ohm)=12W

Therefore, the power supplied to the 1ohm resistor is 4W_, and that to the 3ohm resistor is 12W_

(c)

To determine

The power supplied by each EMFs in the circuit and compare these with the power suppled to each resistors.

(c)

Expert Solution
Check Mark

Answer to Problem 32E

The power supplied by the 6V battery is 12W_, and that by the 2V battery is 4W_. The power supplied to the 1ohm resistor is same as the power supplied by the 2V EMF. The power supplied to the 3ohm resistor is same as the power supplied by the 6V EMF.

Explanation of Solution

Write the expression for the power supplied by the battery (EMF).

  P=IV        (V)

Here, P is the power supplied by the EMF,  V is the potential difference (which is same as the EMF of the battery)

The power supplied to each battery can be computed from equation (V) as,

  P1=IV1P2=IV2}        (VI)

Conclusion:

Substitute 2A for I, 6V for V1, and 2V for V2 in equation (VI) to find the power supplied EMFs.

  P1=(2A)(6V)=12W

  P2=(2A)(2V)=4W

The power supplied to the 1ohm resistor is same as the power supplied by the 2V EMF. The power supplied to the 3ohm resistor is same as the power supplied by the 6V EMF.

Therefore, The power supplied by the 6V battery is 12W_, and that by the 2V battery is 4W_. The power supplied to the 1ohm resistor is same as the power supplied by the 2V EMF. The power supplied to the 3ohm resistor is same as the power supplied by the 6V EMF.

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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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Ohm's law Explained; Author: ALL ABOUT ELECTRONICS;https://www.youtube.com/watch?v=PV8CMZZKrB4;License: Standard YouTube License, CC-BY