Chapter 2, Problem 2.77HP

A voltmeter is used to determine the voltage acrossa resistive element in the circuit of Figure P2.77. The instrument is modeled by an ideal voltmeter in parallel with a $120\text{-k}\Omega$ resistor, as shown. The meter is placedto measure the voltage across $R_4$ . Assume $R_1=8\text{k}\Omega ,R_2=22\text{k}\Omega ,R_3=50\text{k}\Omega ,R_5=125\text{k}\Omega ,$ and is =120 mA. Find the voltage across $R_4$ with and without the voltmeter in the circuit for the following values:
a. $R_4=100\Omega$
b. $R_4=1\Omega$
c. $R_4=10\text{k}\Omega$
d. $R_4=100\text{k}\Omega$

To determine

(a)

The voltage across R₄ with and without voltmeter for the given value.

Answer to Problem 2.77HP

Without the voltmeter

  ${V_{R_4}|}_{R_4=100\Omega }$ = 3.08 V

With the voltmeter:

  ${V_{R_4}|}_{R_4=100\Omega }$ = 3.09 V

Explanation of Solution

Given information:

The given circuit is shown below.

  

  $\begin{array}{l}{R}_1=8\text{k}\Omega \\ R_2=22\text{k}\Omega \\ R_3=50\text{k}\Omega \\ R_5=125\text{k}\Omega \\ R_4=100\Omega \\ i_s=120\text{mA}\end{array}$

Calculation:

Assuming that voltmeter is ideal in parallel with 120-ohm resister.

First develop an expression for V_R4 in terms of R₄ ,using current division

  

Therefore,

  

  $\begin{array}{c}{V}_{R_4}=I_{R_4}{R}_4\\ =I_S\left(\frac{R_S{R}_4}{R_S+R_1+R_2\Vert \left(R_3+R_4\right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+R_4}\right)\\ =\frac{2.129\times {10}^3\cdot R_4}{R_4+68.877\times {10}^3}\end{array}$

Hence,without the voltmeter

  ${V_{R_4}|}_{R_4=100\Omega }$ = 3.08

Now find the voltage drop across $R_4$ with a 120-k( resistor across $R_4$. This is the voltage that the voltmeter will read.

  $I_{R_4}=I_S\left(\frac{R_S}{R_S+R_1+R_2\left|\right|\left(R_3+(R_4||120k\Omega \right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+(R_4||120k\Omega )}\right)$

  $\begin{array}{c}{V}_{R_4}=I_{R_4}{R}_4\\ =I_S\left(\frac{R_S{R}_4}{R_S+R_1+R_2\left|\right|\left(R_3+(R_4||120k\Omega \right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+(R_4||120k\Omega )}\right)\\ =11.272\times {10}^{-3}\cdot \frac{\left(120000+R_4\right)\cdot R_4}{R_4+43.760\times {10}^3}\end{array}$

With the voltmeter:

  ${V_{R_4}|}_{R_4=100\Omega }$ = 3.09 V

To determine

(b)

The voltage across R₄ with and without voltmeter for the given value.

Answer to Problem 2.77HP

Without the voltmeter

  ${V_{R_4}|}_{R_4=1k\Omega }$ = 30.47 V

With the voltmeter:

  ${V_{R_4}|}_{R_4=1k\Omega }$ = 30.47 V

Explanation of Solution

The given circuit is shown below.

  

  $\begin{array}{l}{R}_1=8\text{k}\Omega \\ R_2=22\text{k}\Omega \\ R_3=50\text{k}\Omega \\ R_5=125\text{k}\Omega \\ R_4=1\text{k}\Omega \\ i_s=120\text{mA}\end{array}$

Calculation:

Assuming that voltmeter is ideal in parallel with 120-ohm resister.

First develop an expression for V_R4 in terms of R₄, using current division.

  

Therefore,

  

  $\begin{array}{c}{V}_{R_4}=I_{R_4}{R}_4\\ =I_S\left(\frac{R_S{R}_4}{R_S+R_1+R_2\Vert \left(R_3+R_4\right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+R_4}\right)\\ =\frac{2.129\times {10}^3\cdot R_4}{R_4+68.877\times {10}^3}\end{array}$

Hence, without the voltmeter

  ${V_{R_4}|}_{R_4=1k\Omega }$ = 30.47 V

Now, it is must find that the voltage drop across $R_4$ with a 120-k( resistor across $R_4$. This is the voltage that the voltmeter will read.

  $I_{R_4}=I_S\left(\frac{R_S}{R_S+R_1+R_2\left|\right|\left(R_3+(R_4||120k\Omega \right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+(R_4||120k\Omega )}\right)$

  $\begin{array}{c}{V}_{R_4}=I_{R_4}{R}_4\\ =I_S\left(\frac{R_S{R}_4}{R_S+R_1+R_2\left|\right|\left(R_3+(R_4||120k\Omega \right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+(R_4||120k\Omega )}\right)\\ =11.272\times {10}^{-3}\cdot \frac{\left(120000+R_4\right)\cdot R_4}{R_4+43.760\times {10}^3}\end{array}$

With the voltmeter:

  ${V_{R_4}|}_{R_4=1k\Omega }$ = 30.47 V

To determine

(c)

The voltage across R₄ with and without voltmeter for the given value.

Answer to Problem 2.77HP

Without the voltmeter

  ${V_{R_4}|}_{R_4=10k\Omega }$ = 269.91 V

With the voltmeter:

  ${V_{R_4}|}_{R_4=10k\Omega }$ = 272.58 V

Explanation of Solution

The given circuit is shown below.

  

  $\begin{array}{l}{R}_1=8\text{k}\Omega \\ R_2=22\text{k}\Omega \\ R_3=50\text{k}\Omega \\ R_5=125\text{k}\Omega \\ R_4=10\text{k}\Omega \\ i_s=120\text{mA}\end{array}$

Calculation:

Assuming that voltmeter is ideal in parallel with 120-ohm resister.

First develop an expression for V_R4 in terms of R₄, using current division.

  

Therefore,

  

  $\begin{array}{c}{V}_{R_4}=I_{R_4}{R}_4\\ =I_S\left(\frac{R_S{R}_4}{R_S+R_1+R_2\Vert \left(R_3+R_4\right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+R_4}\right)\\ =\frac{2.129\times {10}^3\cdot R_4}{R_4+68.877\times {10}^3}\end{array}$

Hence, without the voltmeter

  ${V_{R_4}|}_{R_4=10k\Omega }$ = 269.91 V

Now, it is must find that the voltage drop across $R_4$ with a 120-k( resistor across $R_4$. This is the voltage that the voltmeter will read.

  $I_{R_4}=I_S\left(\frac{R_S}{R_S+R_1+R_2\left|\right|\left(R_3+(R_4||120k\Omega \right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+(R_4||120k\Omega )}\right)$

  $\begin{array}{c}{V}_{R_4}=I_{R_4}{R}_4\\ =I_S\left(\frac{R_S{R}_4}{R_S+R_1+R_2\left|\right|\left(R_3+(R_4||120k\Omega \right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+(R_4||120k\Omega )}\right)\\ =11.272\times {10}^{-3}\cdot \frac{\left(120000+R_4\right)\cdot R_4}{R_4+43.760\times {10}^3}\end{array}$

With the voltmeter:

  ${V_{R_4}|}_{R_4=10k\Omega }$ = 272.58 V

To determine

(d)

The voltage across R₄ with and without voltmeter for the given value.

Answer to Problem 2.77HP

Without the voltmeter

  ${V_{R_4}|}_{R_4=100k\Omega }$ = 1260.7 V.

With the voltmeter:

  ${V_{R_4}|}_{R_4=100k\Omega }$ = 1724.99 V.

Explanation of Solution

The given circuit is shown below.

  

  $\begin{array}{l}{R}_1=8\text{k}\Omega \\ R_2=22\text{k}\Omega \\ R_3=50\text{k}\Omega \\ R_5=125\text{k}\Omega \\ R_4=100\text{k}\Omega \\ i_s=120\text{mA}\end{array}$

Calculation:

Assuming that voltmeter is ideal in parallel with 120-ohm resister.

First develop an expression for V_R4 in terms of R₄ ,using current division

  

Therefore

  

  $\begin{array}{c}{V}_{R_4}=I_{R_4}{R}_4\\ =I_S\left(\frac{R_S{R}_4}{R_S+R_1+R_2\Vert \left(R_3+R_4\right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+R_4}\right)\\ =\frac{2.129\times {10}^3\cdot R_4}{R_4+68.877\times {10}^3}\end{array}$

Hence without the voltmeter

  ${V_{R_4}|}_{R_4=100k\Omega }$ = 1260.7 V.

Now it is must find that the voltage drop across $R_4$ with a 120-k( resistor across $R_4$. This is the voltage that the voltmeter will read.

  $I_{R_4}=I_S\left(\frac{R_S}{R_S+R_1+R_2\left|\right|\left(R_3+(R_4||120k\Omega \right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+(R_4||120k\Omega )}\right)$

  $\begin{array}{c}{V}_{R_4}=I_{R_4}{R}_4\\ =I_S\left(\frac{R_S{R}_4}{R_S+R_1+R_2\left|\right|\left(R_3+(R_4||120k\Omega \right)}\right)\cdot \left(\frac{R_2}{R_2+R_3+(R_4||120k\Omega )}\right)\\ =11.272\times {10}^{-3}\cdot \frac{\left(120000+R_4\right)\cdot R_4}{R_4+43.760\times {10}^3}\end{array}$

With the voltmeter:

  ${V_{R_4}|}_{R_4=100k\Omega }$ = 1724.99 V.