Chapter 18, Problem 80P

(a)

To determine

Redraw the circuit to measure the voltage across $83.0\text{ k}\Omega$ resistor.

Answer to Problem 80P

The circuit is redrawn to measure the voltage across $83.0\text{ k}\Omega$ resistor and the circuit is shown below.

Explanation of Solution

The voltmeter should be connected in parallel with the resistor to measure the voltage across it.

Conclusion:

The circuit is shown below,

(b)

To determine

Find the voltage across the $83.0\text{ k}\Omega$ resistor.

Answer to Problem 80P

The voltage across the $83.0\text{ k}\Omega$ resistor is $7.36\text{ mV}$ .

Explanation of Solution

The voltmeter is connected in parallel with the $83.0\text{ k}\Omega$ resistor. Then assume that the voltmeter has zero resistance. The direction of the current are shown in the below figure.

From Kirchhoff’s junction rule,

$I_1=I_2+I_3$ (I)

From Kirchhoff’s loop rule for the left mesh,

 $9.00\text{ V}-I_1\left(35\Omega \right)-I_2\left(1.40\times {10}^3\Omega \right)=0$                                             (II)

From Kirchhoff’s loop rule for the right mesh,

 $I_2\left(1.40\times {10}^3\Omega \right)-I_3\left(16.0\times {10}^3\Omega \right)-I_3\left(83.0\times {10}^3\Omega \right)=0$                      (III)

Conclusion:

Solve equation III to get $I_3$

$\begin{array}{c}{I}_3=\frac{1.40\times {10}^3\Omega }{99.0\times {10}^3\Omega }{I}_2\\ =0.01414I_2\end{array}$ (IV)

Substitute the above equation in equation I.

$\begin{array}{c}{I}_1=I_2+0.01414I_2\\ =1.01414I_2\end{array}$

Substitute the above equation in equation II.

$\begin{array}{c}9.00\text{ V}-1.01414I_2\left(35\Omega \right)-I_2\left(1.40\times {10}^3\Omega \right)=0\\ I_2=\frac{9.00\text{ V}}{1.01414\left(35\Omega \right)+1.40\times {10}^3\Omega }\\ =6.27\text{ mA}\end{array}$

Substitute the above value in equation IV.

$\begin{array}{c}{I}_3=0.01414\left(6.27\text{ mA}\right)\\ =88.7\mu \text{A}\end{array}$

Then, the reading on the voltmeter is,

$\begin{array}{c}V=IR\\ =\left(88.7\mu \text{A}\right)\left(83.0\text{ k}\Omega \right)\\ =7.36\text{ mV}\end{array}$

Therefore, the voltage across the $83.0\text{ k}\Omega$ resistor is $7.36\text{ mV}$.

(c)

To determine

Find the voltage across the $83.0\text{ k}\Omega$ resistor.

Answer to Problem 80P

The voltage across the $83.0\text{ k}\Omega$ resistor is $7.86\text{ mV}$ .

Explanation of Solution

The voltmeter is connected in parallel with the $83.0\text{ k}\Omega$ resistor and the voltmeter has the resistance of $1\text{ M}\Omega$. The direction of the current are shown in the below figure.

From Kirchhoff’s loop rule for the right mesh,

 $I_2\left(1.40\times {10}^3\Omega \right)-I_3\left(16.0\times {10}^3\Omega +{\left(\frac{1}{83.0\times {10}^3\Omega }+\frac{1}{1.00\times {10}^6\Omega }\right)}^{-1}\right)=0$ (V)

Conclusion:

Solve equation V to get $I_3$

$I_2\left(1.40\times {10}^3\Omega \right)-I_3\left(16.0\times {10}^3\Omega +{\left(\frac{1}{83.0\times {10}^3\Omega }+\frac{1}{1.00\times {10}^6\Omega }\right)}^{-1}\right)=0$

$\begin{array}{c}{I}_3=\frac{1.40\times {10}^3\Omega }{16.0\times {10}^3\Omega +{\left(\frac{1}{83.0\times {10}^3\Omega }+\frac{1}{1.00\times {10}^6\Omega }\right)}^{-1}}{I}_2\\ =0.01511I_2\end{array}$ (VI)

Substitute the above equation in equation I.

$\begin{array}{c}{I}_1=I_2+0.01511I_2\\ =1.01511I_2\end{array}$

Substitute the above equation in equation II.

$\begin{array}{c}9.00\text{ V}-1.01511I_2\left(35\Omega \right)-I_2\left(1.40\times {10}^3\Omega \right)=0\\ I_2=\frac{9.00\text{ V}}{1.01511\left(35\Omega \right)+1.40\times {10}^3\Omega }\\ =6.27\text{ mA}\end{array}$

Substitute the above value in equation VI.

$\begin{array}{c}{I}_3=1.01511\left(6.27\text{ mA}\right)\\ =94.7\mu \text{A}\end{array}$

Then, the reading on the voltmeter is,

$\begin{array}{c}V=IR\\ =\left(94.7\mu \text{A}\right)\left(83.0\text{ k}\Omega \right)\\ =7.86\text{ mV}\end{array}$

Therefore, the voltage across the $83.0\text{ k}\Omega$ resistor is $7.86\text{ mV}$.