Question

Question: In Fig. 27-14, assume that $\mathit{\epsilon }\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}{\mathbf{\text{V,r=100Ω,R}}}_{\mathbf{1}}{\mathbf{\text{=250ΩandR}}}_{\mathbf{2}}\mathbf{\text{=}}\mathbf{}\mathbf{300}\mathit{\Omega }\mathbf{,}$ . If the voltmeter resistance R_V= 5. 0 K$\Omega$, what percent error does it introduce into the measurement of the potential difference across R₁ ? Ignore the presence of the ammeter.

Short answer

The error percentage that the voltmeter resistance introduces into the measurement of the potential difference across R₁ is $-0.03\text{or}-3.0\%$.

Step-by-step solution

The given data

$$\begin{gathered} \text{Thegivenemfofthebattery}\epsilon =3\text{V,} \\ {\text{Thegivenresistances,r=100Ω,R}}_1\text{=250}\Omega {\text{,R}}_2\text{=300}\Omega \\ {\text{VoltmeterresistanceR}}_v\text{=5K}\Omega \end{gathered}$$

Understanding the concept of current law

Kirchhoff’s voltage law states that in any closed loop network, the total voltage around the loop is equal to zero. Now, considering each loop, and applying the b=voltage law, we can get the required voltage equation for both loops. Further solving them, we can get the voltmeter reading and the required current value in absence of the ammeter resistance. Using this, the fractional error is calculated.

Formulae:

The voltage equation using Ohm’s law, V = I R (1)

Kirchhoff’s voltage law, $\sum _{\text{closed loop}}\text{V}=0$ (2)

Calculation of the error percentage

The current in R₂ is i . Let be the current in R1 and take it to be the downward direction. According to the junction rule, the current in the voltmeter is found to be i -i₁ and it is downward.

Now, applying equation (1) in equation (2) for the left loop, the voltage equation can be given as:

$\epsilon -i{R}_2-i_1{R}_1-ir=0....................\left(3\right)$

And, applying equation (1) in equation (2) for the right loop, the voltage equation can be given as:

$$\begin{gathered} i_1{R}_1-\left(i-i_1\right)R_V=0 \\ i=\frac{R_1+R_V}{R_V}{i}_1.......................\left(4\right) \end{gathered}$$

Now, substituting the value from equation (4) in equation (3), we get the equation as:

$\epsilon -\frac{{\displaystyle \left(R_2+r\right)\left(R_1+R_V\right)}}{R_V}{i}_1+i_1{R}_1=0$

Thus, the current equation can be given as:

$i_1=\frac{\epsilon R_V}{\left(R_2+r\right)\left(R_1+R_V\right)+R_1{R}_V}$

Now, the voltmeter reading can be given using the given data in the above equation as follows:

$$\begin{gathered} i_1{R}_1=\frac{\epsilon R_V{R}_1}{\left(R_2+r\right)\left(R_1+R_V\right)+R_1{R}_V} \\ =\frac{\left(3.0v\right){\displaystyle \left(5\times {10}^3\Omega \right)}{\displaystyle \left(250\Omega \right)}}{\left(300\Omega +100\Omega \right){\displaystyle \left(250\Omega +5\times {10}^3\Omega \right)}{\displaystyle \left(5\times {10}^3\Omega \right)}{\displaystyle \left(250\Omega \right)}} \\ =1.12V \end{gathered}$$

The current in the absence of the voltmeter can be obtained by taking the limit as becomes infinitely large. Then, the voltmeter reading is given as:

$$\begin{gathered} i{}_1{R}_1=\frac{\epsilon R_1}{\left(R_1+R_2+r\right)} \\ =\frac{\left(3.0v\right){\displaystyle \left(250\Omega \right)}}{\left(250\Omega +300\Omega +100\Omega \right)} \\ ==1.15\text{V} \end{gathered}$$

The fractional error can be given as follows:

$$\begin{gathered} Error=\frac{1.12V-1.15V}{1.15V} \\ =-0.03 \\ =-3.0\% \end{gathered}$$

Hence, the fractional error is $-0.03\text{or}-3.0\%$.