what part of the current is being shunted through the meter?
143/(143+654)
That is the error.
A 65 V. battery is connected across a 65400 ohm and a 37700 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 143000 ohms?
Not sure where to go with this problem, Its easy to get the current for the series but im not sure how the voltmeter resistance factors in.
3 answers
this didn't seem to work. do the voltmeter and resistor have to be treated as a parallel connection?
Given:
E = 65 Volts.
R1 = 65,400 Ohms.
R2 = 37,700Ohms.
Rm = 143,000 Ohms.
Voltmeter disconnected:
I = E/(R1+R2) = 65/(65,400+37,700) = 0.00063A.
V1 = I*R1 = 0.00063 * 65,400 = 41.23 Volts.
Voltmeter connected:
1/R = 1/R1 + 1/Rm = 1/65,400+1/143,000,
R = 44,883.3 Ohms.
I = E/(R+R2) = 65/(44,883.3+37,700) = 0.000787A.
V1 = I*R = 0.000787 * 44,883.3 = 35.33 Volts.
%Error = ((41.23-35.33)/41.23) * 100% = 14.3.
E = 65 Volts.
R1 = 65,400 Ohms.
R2 = 37,700Ohms.
Rm = 143,000 Ohms.
Voltmeter disconnected:
I = E/(R1+R2) = 65/(65,400+37,700) = 0.00063A.
V1 = I*R1 = 0.00063 * 65,400 = 41.23 Volts.
Voltmeter connected:
1/R = 1/R1 + 1/Rm = 1/65,400+1/143,000,
R = 44,883.3 Ohms.
I = E/(R+R2) = 65/(44,883.3+37,700) = 0.000787A.
V1 = I*R = 0.000787 * 44,883.3 = 35.33 Volts.
%Error = ((41.23-35.33)/41.23) * 100% = 14.3.