Question

A voltmeter with internal resistance of 20 kQ (kilo ohms) is connected directly across a resistor (short shunt) and an ammeter is connected in series. The ammeter reads 4 mA (milliamps) and the voltmeter reads 24 Volts.

a) Calculate the resistor value from these readings

b) What is the true value of the resistor

c) What is the percentage error using this method

Answer 1

Question 13624

$$U _ {V} = 2 4 V, I _ {a} = 4 \mathrm {m A}, R _ {V} = 2 0 \mathrm {k} \Omega$$

a) The resistor value can be approximately calculated as $R' = \frac{U_V}{I_a} = 6000\Omega = 6\mathrm{k}\Omega$

b) The true value of resistor is $R = \frac{U_R}{I_R}$ . Since resistor and voltmeter are connected in parallel,

$$I _ {a} = I _ {V} + I _ {R}, U _ {R} = U _ {V}, \text { and } R = \frac {U _ {R}}{I _ {R}} = \frac {U _ {V}}{I _ {a} - I _ {V}} = \frac {U _ {V}}{I _ {a} - \frac {U _ {V}}{R _ {V}}} = \frac {U _ {V} R _ {V}}{I _ {a} \left(R _ {V} - \frac {U _ {V}}{I _ {a}}\right)} = \frac {R ^ {\prime} R _ {V}}{R _ {V} - R ^ {\prime}} \approx 8 5 7 1. 4 3 \Omega$$

c) The percentage error $k = \frac{\Delta R}{R} \cdot 100\% = \frac{R - R'}{R} \cdot 100\% = \frac{8571.43\Omega - 6000\Omega}{8571.43\Omega} \cdot 100\% = 30\%$