Question

The voltage across a resistance R5 in the circuit is to be

measured by a voltmeter connected across it.

(a) If the voltmeter has an internal resistance (Rm) of 4750, what is

the measurement error? (5%)

(b) What value would the voltmeter internal resistance need to be

in order to reduce the measurement error to 1%? (24.750 ohm)

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Answer 1

(a)

The circuit in which the voltage across the resistance R₅ is measured is:

In Figure 1, the resistance R_m acts like a shunt resistance to R₅ and hence, decreases the resistance between the terminals AB. So, the voltage E_m measured by the voltmeter V is not equal to the voltage value E₀.

Thevenin’s theorem is used to find the voltage E₀ in which the circuit in Figure 1 is replaced by the equivalent circuit with a single voltage source E_s and an equivalent resistance R_AB. For this purpose, all the voltage sources E₁ and E₂ are short-circuited.

Draw the circuit in which all the voltage sources are short-circuited for finding the Thevenin equivalent resistance as shown in Figure 2.

Draw the Thevenin equivalent network for the circuit of Figure 1 as shown in Figure 3.

In Figure 2, resistances R₁ and R₂ are in series to each other and the pair of series resistances R₁ and R₂ is in parallel to R₃ across the terminal CD.

…… (1)

Substitute 200Ω for R₁, 300Ω for R₂ and 500Ω for R₃ in equation (1).

In Figure 2, resistances R_CD and R₄ are series to each other and the pair of series resistances R_CD and R₄ is parallel to R₅ across the terminal AB.

…… (2)

Substitute 250Ω for R_CD, 250Ω for R₄ and 500Ω for R₅ in equation (2).

From Figure 3, it is found that the current I flowing through the circuit is:

…… (3)

The voltage measured by the voltmeter E_m is:

…… (4)

Substitute for I in equation (4).

…… (5)

The measurement error of the voltmeter is defined by the term E₀ - E_m

Equation (5) can be rewritten as follows to find the measurement error of the voltmeter.

…… (6)

Substitute 4750Ω for R_m and 250Ω for R_AB in equation (6).

Hence, the measurement error of the voltmeter V is 5%.

(b)

Since the measurement error of the voltmeter is 1%.

Therefore,

E₀ - E_m=0.01E₀

Substitute  0.01E₀ for E₀ - E_m and 250 ? for R_AB in equation (6).

250 + R_m = 25000 

R_m = 25000 - 250 = 24.75kΩ

Hence, the internal resistance of the voltmeter with a measurement error of 1% is 24.75kΩ.