Chapter 23, Problem 91A

(a)

To determine

The output across the resistance R₂ using an ideal voltmeter.

Answer to Problem 91A

The output voltage across R₂ is $6\text{ V}$.

Explanation of Solution

Given:

A voltage divider circuit consists of two resistors of value, $R_1=R_2=47\text{ k}\Omega$ . They are connected to a 12-V battery. An ideal voltmeter is connected across R₂ .

Formula Used:

The voltage across a resistor in a potential divider circuit with a supply voltage V is given by,

$V_{R2}=V\frac{R_2}{R_1+R_2}$

Calculation:

Consider the potential divider circuit shown in Figure 1. Here, the supply voltage is 12-V and the two resistors have 47-kΩ.

Figure 1

In the given case, the voltmeter is ideal and hence has no resistance. Thus, it measures the voltage across $R_2$ as

$\begin{array}{c}{V}_{R2}=V\frac{R_2}{R_1+R_2}\\ =12\times \frac{47}{47+47}\\ =6\text{ V}\end{array}$

Conclusion:

The voltage across R₂ as measured using an ideal voltmeter is 6 V.

(b)

To determine

The output across the resistance R₂ using an ideal voltmeter.

Answer to Problem 91A

The output voltage across R₂is $\text{4}\text{.69 V}$.

Explanation of Solution

Given:

A voltage divider circuit consists of two resistors of value, $R_1=R_2=47\text{ k}\Omega$ is connected to a 12-V battery. A voltmeter of resistance 85 kΩ is connected across R_{2 .}

Formula Used:

The voltage across a resistor in a potential divider circuit with a supply voltage V is given by,

$V_{R2}=V\frac{R_2}{R_1+R_2}$

Calculation:

Consider the potential divider circuit shown in Figure 2. Here, the supply voltage is 12-V and the two resistors have 47-kΩ. The voltmeter connected across R₂ is having a resistance of 85 kΩ.

Figure 2

In the given case, the voltmeter is having a resistance and this appears in parallel with $R_2$ . Thus, the voltmeter measures the voltage across the equivalent of these two resistance values which are given by,

$\begin{array}{c}{R}_{eq}=\frac{R_v{R}_2}{R_v+R_2}\\ =\frac{85\times 47}{85+47}\\ =30.2\text{ k}\Omega \end{array}$

Now, the voltage measured by the voltmeter is given by

$\begin{array}{c}{V}_{R2}=V\frac{R_{eq}}{R_1+R_{eq}}\\ =12\times \frac{30.2}{30.2+47}\\ =4.69\text{ V}\end{array}$

Conclusion:

The voltage across R₂ as measured using a voltmeter of given resistance is 4.69 V.

(b)

To determine

The output across the resistance R₂ .

Answer to Problem 91A

The output voltage across R₂ is $\text{5}\text{.99 V}$.

Explanation of Solution

Given:

A voltage divider circuit consists of two resistors of value, $R_1=R_2=47\text{ k}\Omega$ is connected to a 12-V battery. A voltmeter of resistance 10 MΩ is connected across R₂

Formula Used:

The voltage across a resistor in a potential divider circuit with a supply voltage V is given by,

$V_{R2}=V\frac{R_2}{R_1+R_2}$

Calculation:

Consider the potential divider circuit shown in Figure 2. Here, the supply voltage is 12-V and the two resistors have 47-kΩ. The voltmeter connected across R₂ is having a resistance of 10 M?.

Figure 3

In the given case, the voltmeter is having a resistance and this appears in parallel with $R_2$ . Thus, the voltmeter measures the voltage across the equivalent of these two resistance values which are given by,

$\begin{array}{c}{R}_{eq}=\frac{R_v{R}_2}{R_v+R_2}\\ =\frac{10\times {10}^6\times 47\times {10}^3}{10\times {10}^6+47\times {10}^3}\\ =46.78\text{ k}\Omega \end{array}$

Now, the voltage measured by the voltmeter is given by

$\begin{array}{c}{V}_{R2}=V\frac{R_{eq}}{R_1+R_{eq}}\\ =12\times \frac{46.78}{46.78+47}\\ =5.99\text{ V}\end{array}$

Conclusion:

The voltage across R₂ as measured using a voltmeter of given resistance is 5.99 V. It can be observed that as the resistance of the voltmeter is made higher, the measured voltage value is close to ideal case.