Chapter 2, Problem 67P

1. (a) Obtain the voltage V_o in the circuit of Fig. 2.127(a).
2. (b) Determine the voltage V′_o measured when a voltmeter with 6-kΩ internal resistance is connected as shown in Fig. 2.127(b).
3. (c) The finite resistance of the meter introduces an error into the measurement. Calculate the percent error as

$\left|\frac{V_o-{V^{\prime }}_o}{V_o}\right|\times 100\%$

1. (d) Find the percent error if the internal resistance were 36 kΩ.

Figure 2.127

To determine

Find the value of voltage $V_o$ in Figure 2.127(a).

Answer to Problem 67P

The value of voltage $V_o$ in Figure 2.127(a) is $\underset{\_}{4\text{V}}$.

Explanation of Solution

Calculation:

Refer to Figure 2.127(a) in the textbook.

Consider the current through $4\text{k}\Omega$ resistor as $I_o$.

From current division rule, find the current through $4\text{k}\Omega$ resistor.

$\begin{array}{c}{I}_o=\left(\frac{5\text{k}\Omega }{5\text{k}\Omega +1\text{k}\Omega +4\text{k}\Omega }\right)\left(2\text{mA}\right)\\ =\left(\frac{5\text{k}\Omega }{10\text{k}\Omega }\right)\left(2\text{mA}\right)\\ =\left(\frac{1}{2}\right)\left(2\text{mA}\right)\\ =1\text{mA}\end{array}$

From Ohms law, write the expression to find voltage across $4\text{k}\Omega$ resistor $\left(V_o\right)$.

$V_o=I_o\left(4\text{k}\Omega \right)$

Substitute $1\text{mA}$ for $I_o$ to obtain the value of $V_o$.

$\begin{array}{c}{V}_o=\left(1\text{mA}\right)\left(4\text{k}\Omega \right)\\ =4\text{V}\end{array}$

Conclusion:

Thus, the value of voltage $V_o$ in Figure 2.127(a) is $\underset{\_}{4\text{V}}$.

To determine

Find the voltage ${V^{\prime }}_o$ measured by the voltmeter in Figure 2.127(b).

Answer to Problem 67P

The voltage ${V^{\prime }}_o$ measured by the voltmeter in Figure 2.127(b) is $\underset{\_}{2.856\text{V}}$.

Explanation of Solution

Calculation:

Refer to Figure 2.127(b) in the textbook.

In Figure 2.127(b), as internal resistance of $6\text{k}\Omega$ and $4\text{k}\Omega$ are connected in parallel, therefore the equivalent resistance in parallel connected resistors are calculated as follows.

$\begin{array}{c}{R}_{\text{eq}}=\frac{\left(4\text{k}\Omega \right)\left(6\text{k}\Omega \right)}{4\text{k}\Omega +6\text{k}\Omega }\\ =\frac{24\text{M}\Omega }{10\text{k}\Omega }\\ =2.4\text{k}\Omega \end{array}$

Consider the current through $2.4\text{k}\Omega$ internal resistor as ${I^{\prime }}_o$.

From current division rule, find the current through $2.4\text{k}\Omega$ resistor.

$\begin{array}{c}{I^{\prime }}_o=\left(\frac{5\text{k}\Omega }{2.4\text{k}\Omega +1\text{k}\Omega +5\text{k}\Omega }\right)\left(2\text{mA}\right)\\ =\left(\frac{5\text{k}\Omega }{8.4\text{k}\Omega }\right)\left(2\text{mA}\right)\\ =\left(0.595\right)\left(2\text{mA}\right)\\ =1.19\text{mA}\end{array}$

From Ohms law, write the expression to find voltage across $2.4\text{k}\Omega$ resistor $\left({V^{\prime }}_o\right)$.

${V^{\prime }}_o={I^{\prime }}_o\left(2.4\text{k}\Omega \right)$

Substitute $1.19\text{mA}$ for ${I^{\prime }}_o$ to obtain the value of ${V^{\prime }}_o$.

$\begin{array}{c}{V^{\prime }}_o=\left(1.19\text{mA}\right)\left(2.4\text{k}\Omega \right)\\ =2.856\text{V}\end{array}$

Conclusion:

Thus, the voltage ${V^{\prime }}_o$ measured by the voltmeter in Figure 2.127(b) is $\underset{\_}{2.856\text{V}}$.

To determine

Find the percent error.

Answer to Problem 67P

The error percent is $\underset{\_}{28.6\%}$.

Explanation of Solution

Given data:

The expression for percent error is,

$\text{Percent error}=\left|\frac{V_o-{V^{\prime }}_o}{V_o}\right|\times 100\%$ (1)

Calculation:

Substitute $4\text{V}$ for $V_o$ and $2.856\text{V}$ for ${V^{\prime }}_o$ in equation (1) to obtain the error percent.

$\begin{array}{c}\text{Percent error}=\left|\frac{4\text{V}-2.856\text{V}}{4\text{V}}\right|\times 100\%\\ =\left|\frac{1.144\text{V}}{4\text{V}}\right|\times 100\%\\ =\left|0.286\right|\times 100\%\\ =28.6\%\end{array}$

Conclusion:

Thus, the error percent is $\underset{\_}{28.6\%}$.

To determine

Find the percent error, when internal resistance is $36\text{k}\Omega$

Answer to Problem 67P

The error percentage, when internal resistance is $36\text{k}\Omega$ is $\underset{\_}{6.25\%}$.

Explanation of Solution

Calculation:

Consider internal resistance as $36\text{k}\Omega$. In Figure 2.127(b), as internal resistance of $36\text{k}\Omega$ and $4\text{k}\Omega$ are connected in parallel, therefore the equivalent resistance in parallel connected resistors are calculated as follows.

$\begin{array}{c}{R}_{\text{eq}}=\frac{\left(4\text{k}\Omega \right)\left(36\text{k}\Omega \right)}{4\text{k}\Omega +36\text{k}\Omega }\\ =\frac{144\text{M}\Omega }{40\text{k}\Omega }\\ =3.6\text{k}\Omega \end{array}$

Consider the current through $3.6\text{k}\Omega$ internal resistor as ${I^{\prime }}_o$.

From current division rule, find the current through $3.6\text{k}\Omega$ resistor.

$\begin{array}{c}{I^{\prime }}_o=\left(\frac{5\text{k}\Omega }{3.6\text{k}\Omega +1\text{k}\Omega +5\text{k}\Omega }\right)\left(2\text{mA}\right)\\ =\left(\frac{5\text{k}\Omega }{9.6\text{k}\Omega }\right)\left(2\text{mA}\right)\\ =\left(0.5208\right)\left(2\text{mA}\right)\\ =1.0416\text{mA}\end{array}$

From Ohms law, write the expression to find voltage across $3.6\text{k}\Omega$ resistor $\left({V^{\prime }}_o\right)$.

${V^{\prime }}_o={I^{\prime }}_o\left(3.6\text{k}\Omega \right)$

Substitute $1.0416\text{mA}$ for ${I^{\prime }}_o$ to obtain the value of ${V^{\prime }}_o$.

$\begin{array}{c}{V^{\prime }}_o=\left(1.0416\text{mA}\right)\left(3.6\text{k}\Omega \right)\\ =3.749\text{V}\end{array}$

Substitute $4\text{V}$ for $V_o$ and $3.749\text{V}$ for ${V^{\prime }}_o$ in equation (1) to obtain the error percent.

$\begin{array}{c}\text{Percent error}=\left|\frac{4\text{V}-3.749\text{V}}{4\text{V}}\right|\times 100\%\\ =\left|\frac{0.251\text{V}}{4\text{V}}\right|\times 100\%\\ =\left|0.06275\right|\times 100\%\\ =6.275\%\end{array}$

Conclusion:

Thus, the error percent is $\underset{\_}{6.275\%}$.