Question

For a circuit similar to the one shown in Problem 2-1,$R_1=1.00k\Omega ,R_2=2.00k\Omega ,R_3=4.00k\Omega ,$ and

V = 24.0 V. A voltmeter was placed across contacts 2 and 4. Calculate the relative error in the voltage

reading if the internal resistance of the voltmeter was (a) 4000 $\Omega$, (b) 80.0 k$\Omega$, and (c) 1.00 M$\Omega$.

Short answer

The percentage error is 17.64%, 1.07% and 0.17% respectively.

Step-by-step solution

Given information

Given the resistances$R_1=1.00k\Omega ,R_2=2.00k\Omega ,R_3=4.00k\Omega ,$

and a voltmeter is connected between the points 2 and 4, the diagram is shown below

Part (a)

The internal resistance of the voltmeter is $R_v=4000\Omega$.

Initially, the resistances are connected in series, so the voltage drop across points 2 and 4 is:

$$\begin{gathered} V_{24}=\frac{R_2+R_3}{R_1+R_2+R_3}V \\ {V}_{24}=\frac{2+4}{1+2+4}\times 24V \\ {V}_{24}=20.57V \end{gathered}$$

Now since the voltmeter is connected in parallel at between points 2 and 4. So the equivalent resistance between points 2 and 4 is:

$$\begin{gathered} R\text{'}=\frac{(R_2+R_3)R_v}{R_2+R_3+R_v} \\ R\text{'}=\frac{6000\times 4000}{10000} \\ R\text{'}=2400\Omega \end{gathered}$$

Now the voltage drop across point 2 and 4 is:

$$\begin{gathered} V{\text{'}}_{24}=\frac{2400}{2400+1000}\times 24V \\ V{\text{'}}_{24}=16.94V \end{gathered}$$

Error percentage:

$$\begin{gathered} =\frac{V{\text{'}}_{24}-V_{24}}{V_{24}}\times 100 \\ =\frac{16.94-20.57}{20.57}\times 100 \\ =-17.64\% \end{gathered}$$

Part (b)

The internal resistance of the voltmeter is 80$k\Omega$.

Since the voltmeter is connected in parallel at between points 2 and 4. So the equivalent resistance between points 2 and 4 is:

role="math" localid="1645447912111" $$\begin{gathered} R\text{'}=\frac{(R_2+R_3)R_v}{R_2+R_3+R_v} \\ R\text{'}=\frac{6000\times 4000}{10000} \\ R\text{'}=2400\Omega \end{gathered}$$

Now the voltage drop across point 2 and 4 is:

$$\begin{gathered} V{\text{'}}_{24}=\frac{5581}{5581+1000}\times 24V \\ V{\text{'}}_{24}=20.35V \end{gathered}$$

Error percentage:

$$\begin{gathered} =\frac{V{\text{'}}_{24}-V_{24}}{V_{24}}\times 100 \\ =\frac{20.35-20.57}{20.57}\times 100 \\ =-1.07\% \end{gathered}$$

Part (c)

The internal resistance of the voltmeter is $1M\Omega$.

Since the voltmeter is connected in parallel at between points 2 and 4. So the equivalent resistance between points 2 and 4 is:

$$\begin{gathered} R\text{'}=\frac{(R_2+R_3)R_v}{R_2+R_3+R_v} \\ R\text{'}=\frac{6000\times 100000}{106000} \\ R\text{'}=5660\Omega \end{gathered}$$

Now the voltage drop across point 2 and 4 is:

$$\begin{gathered} V{\text{'}}_{24}=\frac{5660}{5660+1000}\times 24V \\ V{\text{'}}_{24}=20.4V \end{gathered}$$

Error percentage:

$$\begin{gathered} =\frac{V{\text{'}}_{24}-V_{24}}{V_{24}}\times 100 \\ =\frac{20.4-20.57}{20.57}\times 100 \\ =-0.17\% \end{gathered}$$