Chapter 2, Problem 69P

A voltmeter used to measure V_o in the circuit in Fig. 2.129. The voltmeter model consists of an ideal voltmeter in parallel with a 250-kΩ resistor. Let V_s = 95 V, R_s = 25 kΩ, and R₁ = 40 kΩ. Calculate V_o with and without the voltmeter when

1. (a) R₂ = 5 kΩ
2. (b) R₂ = 25 kΩ
3. (c) R₂ = 250 kΩ

Figure 2.129

To determine

Find the values of $V_o$ with and without the voltmeter, when $R_2=5\text{k}\Omega$ in Figure 2.129.

Answer to Problem 69P

The values of $V_o$ with and without the voltmeter, when $R_2=5\text{k}\Omega$ are $\underset{\_}{6.65\text{V}\text{and}\text{6}\text{.783}\text{V}}$, respectively.

Explanation of Solution

Given data:

$R_2$ is $5\text{k}\Omega$.

$V_s$ is $95\text{V}$

$R_s$ is $25\text{k}\Omega$

$R_1$ is $40\text{k}\Omega$

Calculation:

Case 1: with voltmeter

Refer to Figure 2.129 in the textbook.

When a voltmeter is connected to the circuit, $250\text{k}\Omega$ and $R_2$ will be connected in parallel. Therefore the equivalent resistance for the parallel connected circuit is calculated as follows.

$R_{eq}=\frac{R_2\left(250\text{k}\Omega \right)}{R_2+250\text{k}\Omega }$ (1)

By using voltage division rule, write the expression for voltage $V_o$.

$V_o=V_s\left(\frac{R_{eq}}{R_{eq}+R_s+R_1}\right)$ (2)

Substitute $5\text{k}\Omega$ for $R_2$ in equation (1) as follows.

$\begin{array}{c}{R}_{eq}=\frac{\left(5\text{k}\Omega \right)\left(250\text{k}\Omega \right)}{5\text{k}\Omega +250\text{k}\Omega }\\ =\frac{1250\text{M}\Omega }{255\text{k}\Omega }\\ =4.901\text{k}\Omega \end{array}$

Substitute $4.901\text{k}\Omega$ for $R_{eq}$, $25\text{k}\Omega$ for $R_s$, $40\text{k}\Omega$ for $R_1$ and $95\text{V}$ for $V_s$ in equation (1) to obtain the value of $V_o$ with voltmeter.

$\begin{array}{c}{V}_o=\left(95\text{V}\right)\left(\frac{4.901\text{k}\Omega }{4.901\text{k}\Omega +25\text{k}\Omega +40\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(\frac{4.901\text{k}\Omega }{69.901\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(0.07\right)\\ =6.65\text{V}\end{array}$

Case 2: without voltmeter

As there is no voltmeter is connected, by directly using voltage law, write the expression for voltage $V_o$

$V_o=V_s\left(\frac{R_2}{R_s+R_1+R_2}\right)$ (2)

Substitute $5\text{k}\Omega$ for $R_2$, $25\text{k}\Omega$ for $R_s$, $40\text{k}\Omega$ for $R_1$ and $95\text{V}$ for $V_s$ in equation (2) to obtain the value of $V_o$ at without voltmeter.

$\begin{array}{c}{V}_o=\left(95\text{V}\right)\left(\frac{5\text{k}\Omega }{5\text{k}\Omega +25\text{k}\Omega +40\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(\frac{5\text{k}\Omega }{70\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(0.0714\right)\\ =6.783\text{V}\end{array}$

Conclusion:

Thus, the values of $V_o$ with and without the voltmeter, when $R_2=5\text{k}\Omega$ are $\underset{\_}{6.65\text{V}\text{and}\text{6}\text{.783}\text{V}}$ respectively.

To determine

Find the values of $V_o$ with and without the voltmeter, when $R_2=25\text{k}\Omega$ in Figure 2.129.

Answer to Problem 69P

The values of $V_o$ with and without the voltmeter, when $R_2=25\text{k}\Omega$ are $\underset{\_}{24.605\text{V}\text{and}26.38\text{V}}$, respectively.

Explanation of Solution

Given data:

$R_2$ is $25\text{k}\Omega$.

Calculation:

Case 1: with voltmeter

Substitute $25\text{k}\Omega$ for $R_2$ in equation (1) as follows.

$\begin{array}{c}{R}_{eq}=\frac{\left(25\text{k}\Omega \right)\left(250\text{k}\Omega \right)}{25\text{k}\Omega +250\text{k}\Omega }\\ =\frac{6250\text{M}\Omega }{275\text{k}\Omega }\\ =22.727\text{k}\Omega \end{array}$

Substitute $22.727\text{k}\Omega$ for $R_{eq}$, $25\text{k}\Omega$ for $R_s$, $40\text{k}\Omega$ for $R_1$ and $95\text{V}$ for $V_s$ in equation (1) to obtain the value of $V_o$ with voltmeter.

$\begin{array}{c}{V}_o=\left(95\text{V}\right)\left(\frac{22.727\text{k}\Omega }{22.727\text{k}\Omega +25\text{k}\Omega +40\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(\frac{22.727\text{k}\Omega }{87.727\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(0.259\right)\\ =24.605\text{V}\end{array}$

Case 2: without voltmeter

Substitute $25\text{k}\Omega$ for $R_2$, $25\text{k}\Omega$ for $R_s$, $40\text{k}\Omega$ for $R_1$ and $95\text{V}$ for $V_s$ in equation (2) to obtain the value of $V_o$ at without voltmeter.

$\begin{array}{c}{V}_o=\left(95\text{V}\right)\left(\frac{25\text{k}\Omega }{25\text{k}\Omega +25\text{k}\Omega +40\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(\frac{25\text{k}\Omega }{90\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(0.2777\right)\\ =26.38\text{V}\end{array}$

Conclusion:

Thus, the values of $V_o$ with and without the voltmeter, when $R_2=25\text{k}\Omega$ are $\underset{\_}{24.605\text{V}\text{and}26.38\text{V}}$, respectively.

To determine

Find the values of $V_o$ with and without the voltmeter, when $R_2=25\text{k}\Omega$ in Figure 2.129.

Answer to Problem 69P

The values of $V_o$ with and without the voltmeter, when $R_2=250\text{k}\Omega$ are $\underset{\_}{62.41\text{V}\text{and}75.335\text{V}}$, respectively.

Explanation of Solution

Given data:

$R_2$ is $250\text{k}\Omega$.

Calculation:

Case 1: with voltmeter

Substitute $250\text{k}\Omega$ for $R_2$ in equation (1) as follows.

$\begin{array}{c}{R}_{eq}=\frac{\left(250\text{k}\Omega \right)\left(250\text{k}\Omega \right)}{250\text{k}\Omega +250\text{k}\Omega }\\ =\frac{62500\text{M}\Omega }{500\text{k}\Omega }\\ =125\text{k}\Omega \end{array}$

Substitute $125\text{k}\Omega$ for $R_{eq}$, $25\text{k}\Omega$ for $R_s$, $40\text{k}\Omega$ for $R_1$ and $95\text{V}$ for $V_s$ in equation (1) to obtain the value of $V_o$ with voltmeter.

$\begin{array}{c}{V}_o=\left(95\text{V}\right)\left(\frac{125\text{k}\Omega }{125\text{k}\Omega +25\text{k}\Omega +40\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(\frac{125\text{k}\Omega }{190\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(0.657\right)\\ =62.41\text{V}\end{array}$

Case 2: without voltmeter

Substitute $250\text{k}\Omega$ for $R_2$, $25\text{k}\Omega$ for $R_s$, $40\text{k}\Omega$ for $R_1$ and $95\text{V}$ for $V_s$ in equation (2) to obtain the value of $V_o$ at without voltmeter.

$\begin{array}{c}{V}_o=\left(95\text{V}\right)\left(\frac{250\text{k}\Omega }{250\text{k}\Omega +25\text{k}\Omega +40\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(\frac{250\text{k}\Omega }{315\text{k}\Omega }\right)\\ =\left(95\text{V}\right)\left(0.793\right)\\ =75.335\text{V}\end{array}$

Conclusion:

Thus, the values of $V_o$ with and without the voltmeter, when $R_2=250\text{k}\Omega$ are $\underset{\_}{62.41\text{V}\text{and}75.335\text{V}}$, respectively.