Chapter 18, Problem 127P

To determine

Find the voltmeter reading.

Answer to Problem 127P

The voltmeter reading is $7.22\text{ V}$.

Explanation of Solution

The circuit diagram is shown below,

From the circuit the voltmeter reading is,

$V_4=I_4\left(83.0\Omega \right)$ (I)

From Kirchhoff’s junction rule,

$I_1=I_2+I_3$ (II)

$I_3=I_4+I_5$ (III)

Write the equation to find $I_1$ .

$I_1=\frac{\epsilon }{R_{eq}}$ (IV)

Here, $I_1$ is the current, $\epsilon$ is potential difference and $R_{eq}$ is the equivalent resistance.

From the circuit, consider $35\Omega$ resistor as $R_1$, $\text{1}\text{.40 k}\Omega$ resistor as $R_2$, $16\text{ k}\Omega$ resistor as $R_3$, $83\text{ k}\Omega$ resistor as $R_4$ and voltmeter resistance $670\text{ k}\Omega$ as $R_5$.

The resistors $R_4$ and $R_5$ are in parallel then the equivalent resistance is,

$\begin{array}{l}\frac{1}{R\text{'}}=\frac{1}{R_4}+\frac{1}{R_5}\\ R\text{'}={\left(\frac{1}{R_4}+\frac{1}{R_5}\right)}^{-1}\end{array}$

This resistance is in series with $R_3$ , the equivalent resistance is,

$R\text{'}\text{'}=R_3+R\text{'}$

This resistance is in parallel with $R_2$ then the equivalent resistance is,

$\begin{array}{l}\frac{1}{R\text{'}\text{'}\text{'}}=\frac{1}{R_2}+\frac{1}{R\text{'}\text{'}}\\ R\text{'}\text{'}\text{'}=\frac{1}{\frac{1}{R_2}+\frac{1}{R\text{'}\text{'}}}\end{array}$

This resistance is in series with $R_1$, then the final equivalent resistance is,

$R_{eq}=R_1+R\text{'}\text{'}\text{'}$

Substitute the value of $R\text{'}$, $R\text{'}\text{'}$ and $R\text{'}\text{'}\text{'}$ in the above equation.

$R_{eq}=R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3+{\left(\frac{1}{R_4}+\frac{1}{R_5}\right)}^{-1}}}$ (V)

Conclusion:

Substitute $35\Omega$ for $R_1$, $\text{1}\text{.40 k}\Omega$ for $R_2$, $16\text{ k}\Omega$ for $R_3$, $83\text{ k}\Omega$ for $R_4$ and $670\text{ k}\Omega$ for $R_5$ in equation V.

$\begin{array}{c}{R}_{eq}=35\Omega +\frac{1}{\frac{1}{\text{1}\text{.40 k}\Omega }+\frac{1}{16\text{ k}\Omega +{\left(\frac{1}{83\text{ k}\Omega }+\frac{1}{670\text{ k}\Omega }\right)}^{-1}}}\\ =35\Omega +\frac{1}{\frac{1}{\text{1}\text{.40 k}\Omega }+\frac{1}{16\text{ k}\Omega +73851\Omega }}\\ =35\Omega +\frac{1}{\frac{1}{\text{1}\text{.40 k}\Omega }+\frac{1}{89851\Omega }}\\ =35\Omega +1379\Omega \end{array}$

Substitute $35\Omega +1379\Omega$ for $R_{eq}$ and $9.00\text{ V}$ for $\epsilon$ in equation IV.

$\begin{array}{c}{I}_1=\frac{9.00\text{ V}}{35\Omega +1379\Omega }\\ =6.367\times {10}^{-3}\text{ A}\end{array}$

Then the voltage across the all the resistors drawn vertical in the circuit with equivalent resistance of $1379\Omega$ is $\left(6.367\times {10}^{-3}\text{ A}\right)\left(1379\Omega \right)=8.780\text{ V}$.

This voltage is across the two resistors in right side with equivalent resistance $89851\Omega$ and the voltmeter meter. Then the current $I_3$ is $\raisebox{1ex}{$8.780\text{ V}$}\!\left/ \!\raisebox{-1ex}{$89851\Omega $}\right.=97.72\text{ μA}$.

The voltage across the voltmeter and the resistor it is supposed to be measured is then $\left(97.72\text{ μA}\right)\left(73851\Omega \right)=7.22\text{ V}$.

Therefore, the voltmeter reading is $7.22\text{ V}$.