Chapter 19, Problem 58P

To determine

The percent inaccuracy due to meter resistance for each case.

Answer to Problem 58P

The percentage inaccuracy in the voltage across first resistor is $14.32\%$ and the percentage inaccuracy in the voltage across second resistor is $14.28\%$ .

Explanation of Solution

Given info:

The emf of the battery is, $V=45\text{V}$ .

The resistance of the first resistor of the series combination is, $R_1=38\text{kΩ}=38\times {10}^3\text{Ω}$ .

The resistance of the second resistor of the series combination is, $R_2=27\text{kΩ}=27\times {10}^3\text{Ω}$ .

The internal resistance of the voltmeter is, $r=95\text{kΩ}=95\times {10}^3\text{Ω}$ .

Formula Used:

The expression to calculate the voltage across first resistor is,

  $V_1=\frac{V{R}_1}{R_1+R_2}$ …… (1)

The expression to calculate the voltage across second resistor is,

  $V_2=\frac{V{R}_2}{R_1+R_2}$ …… (2)

The expression to calculate the voltage across first resistor by taking internal resistance is,

  $E_1=\frac{V\left(\frac{r\cdot R_1}{r+R_1}\right)}{\left(\frac{r\cdot R_1}{r+R_1}+R_2\right)}$ …… (3)

The expression to calculate the voltage across second resistor by taking internal resistance is,

  $E_2=\frac{V\left(\frac{r\cdot R_2}{r+R_2}\right)}{\left(\frac{r\cdot R_2}{r+R_2}+R_1\right)}$ …… (4)

The expression to calculate the percentage inaccuracy in the voltage across first resistor is,

  $x_1=\left(\frac{V_1-E_1}{V_1}\right)\times 100$ …… (5)

The expression to calculate the percentage inaccuracy in the voltage across second resistor is,

  $x_2=\left(\frac{V_2-E_2}{V_2}\right)\times 100$ …… (6)

Calculation:

Substitute all the values in the expression (1).

  $\begin{array}{c}{V}_1=\frac{\left(45\text{V}\right)\left(38\times {10}^3\text{Ω}\right)}{\left(38\times {10}^3\text{Ω}+27\times {10}^3\text{Ω}\right)}\\ =26.31\text{V}\end{array}$

Substitute all the values in the expression (2).

  $\begin{array}{c}{V}_2=\frac{\left(45\text{V}\right)\left(27\times {10}^3\text{Ω}\right)}{\left(38\times {10}^3\text{Ω}+27\times {10}^3\text{Ω}\right)}\\ =18.69\text{V}\end{array}$

Substitute all the values in the expression (3).

  $\begin{array}{c}{E}_1=\frac{\left(45\text{V}\right)\left(\frac{\left(38\times {10}^3\text{Ω}\right)\left(95\times {10}^3\text{Ω}\right)}{38\times {10}^3\text{Ω}+95\times {10}^3\text{Ω}}\right)}{\left(\frac{\left(38\times {10}^3\text{Ω}\right)\left(95\times {10}^3\text{Ω}\right)}{38\times {10}^3\text{Ω}+95\times {10}^3\text{Ω}}+27\times {10}^3\text{Ω}\right)}\\ =22.54\text{V}\end{array}$

Substitute all the values in the expression (4).

  $\begin{array}{c}{E}_2=\frac{\left(45\text{V}\right)\left(\frac{\left(27\times {10}^3\text{Ω}\right)\left(95\times {10}^3\text{Ω}\right)}{27\times {10}^3\text{Ω}+95\times {10}^3\text{Ω}}\right)}{\left(\frac{\left(27\times {10}^3\text{Ω}\right)\left(95\times {10}^3\text{Ω}\right)}{27\times {10}^3\text{Ω}+95\times {10}^3\text{Ω}}+38\times {10}^3\text{Ω}\right)}\\ =16.02\text{V}\end{array}$

Substitute all the values in the expression (5).

  $\begin{array}{c}{x}_1=\left(\frac{26.31\text{V}-22.54\text{V}}{26.31\text{V}}\right)\times 100\\ =14.32\%\end{array}$

Substitute all the values in the expression (6).

  $\begin{array}{c}{x}_2=\left(\frac{18.69\text{V}-16.02\text{V}}{18.69\text{V}}\right)\times 100\\ =14.28\%\end{array}$

Conclusion:

Thus, the percentage inaccuracy in the voltage across first resistor is $14.32\%$ and the percentage inaccuracy in the voltage across second resistor is $14.28\%$ .