EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 25, Problem 91P

(a)

To determine

The reading on the voltmeter when R is 1.00kΩ .

(a)

Expert Solution
Check Mark

Answer to Problem 91P

The reading of the voltmeter is 3.33V .

Explanation of Solution

Given:

The resistance R is 1.00kΩ .

The resistance RV of the voltmeter is 10MΩ .

The given diagram is shown in Figure 1

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 25, Problem 91P

Figure 1

Formula:

The expression to determine the equivalent resistance of the circuit is given by,

  Req=RvRRv+R=10MΩR10MΩ+R

The expression to determine the value of the current in the circuit is given by,

  I=10VReq+2R

The expression for the voltage reading of the voltmeter is given by,

  V=( 10V R eq +2R)Req=10V1+ 2R R eq =10V1+ 2R 10MΩR 10MΩ+R =( 10V)( 5MΩ)R+15MΩ

Calculation:

The reading of the voltmeter is calculated as,

  V=( 10V)( 5MΩ)R+15MΩ=( 10V)( 5MΩ)( 10 3 kΩ 1MΩ )1kΩ+15MΩ( 10 3 kΩ 1MΩ )=3.33V

Conclusion:

Therefore, the reading of the voltmeter is 3.33V .

(b)

To determine

The reading of the voltmeter.

(b)

Expert Solution
Check Mark

Answer to Problem 91P

The reading of the voltmeter is 3.33V .

Explanation of Solution

Given:

The resistance R is 10kΩ .

Formula:

The expression for the voltage reading of the voltmeter is given by,

  V=(10V)(5MΩ)R+15MΩ

Calculation:

The reading of the voltmeter is calculated as,

  V=( 10V)( 5MΩ)R+15MΩ=( 10V)( 5MΩ)( 10 3 kΩ 1MΩ )10kΩ+15MΩ( 10 3 kΩ 1MΩ )=3.13V

Conclusion:

Therefore, the reading of the voltmeter is 3.33V .

(c)

To determine

The reading of the voltmeter.

(c)

Expert Solution
Check Mark

Answer to Problem 91P

The reading of the voltmeter is 3.13V .

Explanation of Solution

Given:

The resistance R is 1MΩ .

Formula:

The expression for the voltage reading of the voltmeter is given by,

  V=(10V)(5MΩ)R+15MΩ

Calculation:

The reading of the voltmeter is calculated as,

  V=( 10V)( 5MΩ)R+15MΩ=( 10V)( 5MΩ)1MΩ+15MΩ=3.13V

Conclusion:

Therefore, the reading of the voltmeter is 3.13V .

(d)

To determine

The reading of the voltmeter.

(d)

Expert Solution
Check Mark

Answer to Problem 91P

The reading of the voltmeter is 2V .

Explanation of Solution

Given:

The resistance R is 10MΩ .

Formula:

The expression for the voltage reading of the voltmeter is given by,

  V=(10V)(5MΩ)R+15MΩ

Calculation:

The reading of the voltmeter is calculated as,

  V=( 10V)( 5MΩ)R+15MΩ=( 10V)( 5MΩ)10MΩ+15MΩ=2V

Conclusion:

Therefore, the reading of the voltmeter is 2V .

(e)

To determine

The reading of the voltmeter.

(e)

Expert Solution
Check Mark

Answer to Problem 91P

The reading of the voltmeter is 0.435V .

Explanation of Solution

Given:

The resistance R is 100MΩ .

Formula:

The expression for the voltage reading of the voltmeter is given by,

  V=(10V)(5MΩ)R+15MΩ

Calculation:

The reading of the voltmeter is calculated as,

  V=( 10V)( 5MΩ)R+15MΩ=( 10V)( 5MΩ)100MΩ+15MΩ=0.435V

Conclusion:

Therefore, the reading of the voltmeter is 0.435V .

(f)

To determine

The maximum possible value of the resistance R when the measured voltage is 10% of the true voltage.

(f)

Expert Solution
Check Mark

Answer to Problem 91P

The maximum value of R is 1.67MΩ .

Explanation of Solution

Given:

The measured voltage is to be within ten percent of the true voltage.

Formula:

The condition for the voltage is given by,

  VtrueVVtrue=1VVtrue<0.1

The expression for the current I is given by,

  I=10V3R

The expression for the true voltage is given by,

  Vtrue=IR

Calculation:

The expression to determine the resistance is evaluated as,

  1VV true<0.11 ( 10V )( 5MΩ ) R+15MΩIR<0.11 ( 10V )( 5MΩ ) R+15MΩ( 10V 3R )R<0.1R<1.67MΩ

Conclusion:

Therefore, the maximum value of R is 1.67MΩ .

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Chapter 25 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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