Question

In the figure assume that epsilon = 3.5 V, r = 140 ohm, R_1 = 340 ohm, and R_2 = 250 ohm. If the voltmeter resistance is R_v = 5.1 k ohm, what percent error (including sign) does it introduce into the measurement of the potential difference across R_1? Ignore the presence of the ammeter. Number Units

Answer 1

case a) without voltmeter

current in the circuit , i = E /(r + R1 + R2 ) = 3.5/(140+340+250) = 4.79 mA

so voltage across R1 , V1 = 4.79*10^-3*340 = 1.628 volt

case b) with voltmeter

Rv = 5100 ohm

here Rv , R1 are in parallel will equivallent to Req = R1*Rv/(R1+Rv ) = 340*5100/(340+5100) = 318.75 ohm
now current in the circuit , i = 3.5/(140+250+318.75) = 4.93 mA

so voltage across R1 is V1' = Req*3.5/(Req+R2) = 318.75*3.5/(318.75+ 250) = 1.96 volt

SO %ERROR = (V1' - V1)/V1 = (1.96 - 1.628)*100/1.628 = 20.39%