Question

Two resistors of values 1 kΩ and 4 Ω are connected in series across a constant voltage supply of 100 V. A voltmeter having an internal resistance of 12 kΩ is connected across the 4 kΩ resistor. Draw the circuit and calculate 

(a) true voltage across 4 kΩ resistor before the voltmeter was connected. 

(b) actual voltage across 4kΩ resistor after the voltmeter is connected and the voltage recorded by the voltmeter. 

(c) change in supply current when voltmeter is connected. 

(d) percentage error in voltage across 4 kΩ resistor

Answer 1

(a) True voltage drop across 4 kΩ resistor as found by voltage-divider rule is 100 × 4/5 = 80V Current from the supply = 100/(4 + 1) = 20mA 

(b) In Fig. 1.56, voltmeter has been joined across the 4 kΩ resistor. The equivalent resistance between B and C = 4 × 12/16 = 3 kΩ

Drop across B and C = 100 × 3/(3 + 1) = 75 V. 

(c) Resistance between A and C = 3 + 1 = 4 kΩ 

New supply current = 100/4 = 25 mA 

∴ increase in current = 25 − 20 = 5 mA

(d) Percentage error in voltage = (actual voltage true voltage)/(true voltage) = (75 - 80)/80 x 100 = - 6.25%

The reduction in the value of voltage being measured in called voltmeter loading effect because voltmeter loads down the circuit element across which it is connected. Smaller the voltmeter resistance as compared to the resistance across which it is connected, greater the loading effect and, hence, greater the error in the voltage reading. Loading effect cannot be avoided but can be minimized by selecting a voltmeter of resistance much greater than that of the network across which it is connected.