Chapter 25, Problem 91P

(a)

To determine

The reading on the voltmeter when $R$ is $1.00\text{k}\Omega$ .

Answer to Problem 91P

The reading of the voltmeter is $3.33\text{V}$ .

Explanation of Solution

Given:

The resistance $R$ is $1.00\text{k}\Omega$ .

The resistance $R_V$ of the voltmeter is $10\text{M}\Omega$ .

The given diagram is shown in Figure 1

  

Figure 1

Formula:

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

  $\begin{array}{c}{R}_{eq}=\frac{R_{v}R}{R_v+R}\\ =\frac{10\text{M}\Omega R}{10\text{M}\Omega +R}\end{array}$

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

  $I=\frac{10\text{V}}{R_{eq}+2R}$

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

  $\begin{array}{c}V=\left(\frac{10\text{V}}{R_{eq}+2R}\right)R_{eq}\\ =\frac{10\text{V}}{1+\frac{2R}{R_{eq}}}\\ =\frac{10\text{V}}{1+\frac{2R}{\frac{10\text{M}\Omega R}{10\text{M}\Omega +R}}}\\ =\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }\end{array}$

Calculation:

The reading of the voltmeter is calculated as,

  $\begin{array}{c}V=\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }\\ =\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)\left(\frac{{10}^3\text{k}\Omega }{1\text{M}\Omega }\right)}{1\text{k}\Omega +15\text{M}\Omega \left(\frac{{10}^3\text{k}\Omega }{1\text{M}\Omega }\right)}\\ =3.33\text{V}\end{array}$

Conclusion:

Therefore, the reading of the voltmeter is $3.33\text{V}$ .

(b)

To determine

The reading of the voltmeter.

Answer to Problem 91P

The reading of the voltmeter is $3.33\text{V}$ .

Explanation of Solution

Given:

The resistance $R$ is $10\text{k}\Omega$ .

Formula:

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

  $V=\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }$

Calculation:

The reading of the voltmeter is calculated as,

  $\begin{array}{c}V=\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }\\ =\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)\left(\frac{{10}^3\text{k}\Omega }{1\text{M}\Omega }\right)}{10\text{k}\Omega +15\text{M}\Omega \left(\frac{{10}^3\text{k}\Omega }{1\text{M}\Omega }\right)}\\ =3.13\text{V}\end{array}$

Conclusion:

Therefore, the reading of the voltmeter is $3.33\text{V}$ .

(c)

To determine

The reading of the voltmeter.

Answer to Problem 91P

The reading of the voltmeter is $3.13\text{V}$ .

Explanation of Solution

Given:

The resistance $R$ is $1\text{M}\Omega$ .

Formula:

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

  $V=\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }$

Calculation:

The reading of the voltmeter is calculated as,

  $\begin{array}{c}V=\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }\\ =\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{1\text{M}\Omega +15\text{M}\Omega }\\ =3.13\text{V}\end{array}$

Conclusion:

Therefore, the reading of the voltmeter is $3.13\text{V}$ .

(d)

To determine

The reading of the voltmeter.

Answer to Problem 91P

The reading of the voltmeter is $2\text{V}$ .

Explanation of Solution

Given:

The resistance $R$ is $10\text{M}\Omega$ .

Formula:

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

  $V=\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }$

Calculation:

The reading of the voltmeter is calculated as,

  $\begin{array}{c}V=\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }\\ =\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{10\text{M}\Omega +15\text{M}\Omega }\\ =2\text{V}\end{array}$

Conclusion:

Therefore, the reading of the voltmeter is $2\text{V}$ .

(e)

To determine

The reading of the voltmeter.

Answer to Problem 91P

The reading of the voltmeter is $0.435\text{V}$ .

Explanation of Solution

Given:

The resistance $R$ is $100\text{M}\Omega$ .

Formula:

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

  $V=\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }$

Calculation:

The reading of the voltmeter is calculated as,

  $\begin{array}{c}V=\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }\\ =\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{100\text{M}\Omega +15\text{M}\Omega }\\ =0.435\text{V}\end{array}$

Conclusion:

Therefore, the reading of the voltmeter is $0.435\text{V}$ .

(f)

To determine

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

Answer to Problem 91P

The maximum value of $R$ is $1.67\text{M}\Omega$ .

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,

  $\frac{V_{true}-V}{V_{true}}=1-\frac{V}{V_{true}}<0.1$

The expression for the current $I$ is given by,

  $I=\frac{10\text{V}}{3R}$

The expression for the true voltage is given by,

  $V_{true}=IR$

Calculation:

The expression to determine the resistance is evaluated as,

  $\begin{array}{l}1-\frac{V}{V_{true}}<0.1\\ 1-\frac{\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }}{IR}<0.1\\ 1-\frac{\frac{\left(10\text{V}\right)\left(5\text{M}\Omega \right)}{R+15\text{M}\Omega }}{\left(\frac{10\text{V}}{3R}\right)R}<0.1\\ R<1.67\text{M}\Omega \end{array}$

Conclusion:

Therefore, the maximum value of $R$ is $1.67\text{M}\Omega$ .