Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 18, Problem 79P

(a)

To determine

Redraw the circuit to measure the current through 1.40 kΩ resistor.

(a)

Expert Solution
Check Mark

Answer to Problem 79P

The circuit is redrawn to measure the current through 1.40 kΩ resistor and the circuit is shown below.

Explanation of Solution

The ammeters should be connected in series with the resistor to measure the current flowing thought it.

Conclusion:

The circuit is shown below,

Physics, Chapter 18, Problem 79P , additional homework tip  1

(b)

To determine

Find the current through the 1.40 kΩ resistor.

(b)

Expert Solution
Check Mark

Answer to Problem 79P

The current through the 1.40 kΩ resistor is 6.27 mA .

Explanation of Solution

The ammeter is connected in series with the 1.40 kΩ resistor. Then assume that the ammeter resistance is zero. The direction of the current are shown in the below figure.

Physics, Chapter 18, Problem 79P , additional homework tip  2

From Kirchhoff’s junction rule,

I1=I2+I3 (I)

From Kirchhoff’s loop rule for the left mesh,

 9.00 VI1(35 Ω)I2(1.40×103 Ω)=0                                             (II)

From Kirchhoff’s loop rule for the right mesh,

 I2(1.40×103 Ω)I3(16.0×103 Ω)I3(83.0×103 Ω)=0                      (III)

Conclusion:

Solve equation III to get I3

I3=1.40×103 Ω99.0×103 ΩI2=0.01414I2

Substitute the above equation in equation I.

I1=I2+0.01414I2=1.01414I2

Substitute the above equation in equation II.

9.00 V1.01414I2(35 Ω)I2(1.40×103 Ω)=0I2=9.00 V1.01414(35 Ω)+1.40×103 Ω=6.27 mA

Therefore, the current through the 1.40 kΩ resistor is 6.27 mA.

(c)

To determine

Find the current through the 1.40 kΩ resistor.

(c)

Expert Solution
Check Mark

Answer to Problem 79P

The current through the 1.40 kΩ resistor is 5.79 mA .

Explanation of Solution

The ammeter is connected in series with the 1.40 kΩ resistor and the ammeter has the resistance of 120 Ω. The direction of the current are shown in the below figure.

Physics, Chapter 18, Problem 79P , additional homework tip  3

From Kirchhoff’s loop rule for the left mesh,

 9.00 VI1(35 Ω)I2(1.40×103 Ω+120 Ω)=0 (IV)

From Kirchhoff’s loop rule for the right mesh,

 I2(1.40×103 Ω+120 Ω)I3(16.0×103 Ω)I3(83.0×103 Ω)=0                   (V)

Conclusion:

Solve equation V to get I3

I3=1.40×103 Ω+120 Ω99.0×103 ΩI2=0.01535I2

Substitute the above equation in equation I.

I1=I2+0.01535I2=1.01535I2

Substitute the above equation in equation IV.

9.00 V1.01535I2(35 Ω)I2(1.40×103 Ω+120 Ω)=0I2=9.00 V1.01535(35 Ω)+1.40×103 Ω+120 Ω=5.79 mA

Therefore, the current through the 1.40 kΩ resistor is 5.79 mA.

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Chapter 18 Solutions

Physics

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