Question

A 66 V. battery is connected across a 64700 ohm and a 32800 ohm resistor connected in series. What is the per cent error in the voltmeter reading when placed across the first resistor, if the voltmeter resistance is 125000 ohms?

Answer 1

let R1 = 64700 ohm
R2 = 32800 ohm
Rv = 125000 ohm

when voltmeter is not connected,

Req = R1 + R2

= 64700 + 32800

= 97500 ohms

I = V/Req

= 60/97500

= 0.00061538 A

Actual voltage drop across R1 when voltmetere is not connected,

V1 = I*R1

= 0.00061538*64700

= 39.814 V

when voltmeter connected

Req = R1*Rv/(R1 + Rv) + R2

= 64700*125000/(64700 + 125000) + 32800

= 75433 ohm

I = V/Req

= 60/75433

= 0.0007954 A

voltage across R1, V1' = I*R1*Rv/(R1 + Rv)

= 0.0007954*64700*125000/(64700 + 125000)

= 33.91 V

percentage error = (V1 - V1')*100/V1

= (39.814 - 33.91)*100/39.814

= 14.8 % <<<<<<<<----------------Answer