How much does the digital voltmeter affect the voltage being measured for each circuit indicated in Figure 7-36? Assume the meter has an input resistance (RM) of 10 MΩ.

Question

How much does the digital voltmeter affect the voltage being measured for each circuit indicated in Figure 7-36? Assume the meter has an input resistance ([latex]R_{M}[/latex]) of 10 MΩ.

Accepted Answer

To show the small differences more clearly, the results are expressed in more than three significant figures in this example.
(a) Refer to Figure 7-36(a). The unloaded voltage across [latex]R_{2}[/latex] he voltage divider circuit is

[latex]V_{2}= \left(\frac{R_{2}}{R_{1}+ R_{2}} \right)V_{S}= \left(\frac{100\ \Omega }{280\ \Omega } \right) 15 V = 5.357 V[/latex]

The meter's resistance in parallel with [latex]R_{2}[/latex] is

[latex]R_{2}|| R_{M}= \left(\frac{R_{2}R_{M}}{R_{2}+ R_{M}} \right)= \frac{(100 \Omega )(10 M\Omega) }{10.0001 M \Omega} = 99.999 \Omega[/latex]

The voltage actually measured by the meter is

[latex]V_{2}= \left(\frac{R_{2}|| R_{M}}{R_{1}+ R_{2} || R_{M}} \right)V_{S}= \left(\frac{99.999 \Omega }{279.999 \Omega } \right) 15 V = 5.357 V[/latex]

The voltmeter has no measurable loading effect,

(b) Refer to Figure 7-36(b).

[latex]V_{2}= \left(\frac{R_{2}}{R_{1}+ R_{2}} \right)V_{S}= \left(\frac{100k \Omega }{280 k \Omega } \right) 15 V = 5.357 V[/latex]

[latex]R_{2} \parallel R_{M}= \frac{R_{2}R_{M}}{R_{2}+ R_{M}}= \frac{(100 k \Omega)(10M \Omega)}{10.1M \Omega } = 99.01k \Omega[/latex]

The voltage actually measured by the meter is

[latex]V_{2}= \left(\frac{R_{2}|| R_{M}}{R_{1}+ R_{2} || R_{M}} \right)V_{S}= \left(\frac{99.01 k \Omega }{279.01 k \Omega } \right) 15 V = 5.323 V[/latex]

The loading effect of the voltmeter reduces the voltage by a very small amount.

(e) Refer lo Figure 7-36(c).

[latex]V_{2}= \left(\frac{R_{2}}{R_{1}+ R_{2} } \right)V_{S}= \left(\frac{1.0 M \Omega }{2.8 M \Omega } \right) 15 V = 5.357 V[/latex]

[latex]R_{2} \parallel R_{M}= \frac{R_{2}R_{M}}{R_{2}+ R_{M}}= \frac{(1.0 M \Omega)(10 M \Omega)}{11 M \Omega } = 909.09 k \Omega[/latex]

The voltage actually measured is

[latex]V_{2}= \left(\frac{R_{2}|| R_{M}}{R_{1}+ R_{2} || R_{M}} \right)V_{S}= \left(\frac{909.09 k\Omega }{2.709 M\Omega } \right) 15 V = 5.034 V[/latex]

The loading effect of the voltmeter reduces the voltage by a noticeable amount. As you can see, the higher the resistance across which a voltage is measured, the more the loading effect.

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