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Measuring Instruments: Voltmeter
You can measure a voltage by placing a galvanometer in
parallel with the circuit component across which you wish to
measure the potential difference.
RG
G
Vab=?
a
R=10 
r=0.5 
V=3 V
b
Example: an galvanometer of resistance 60  is used to
measure the voltage drop across a 10 k resistor in series with
a 6 V battery and a 5 k resistor (neglect the internal
resistance of the battery). What is the percent error caused by
the nonzero resistance of the galvanometer?
First calculate the actual voltage drop. a
R1=10 k
R eq  R1 +R 2 =15 103 
V
6V
-3
I


0.4

10
A
3
R eq 15 10 
Vab = IR   0.4 10-3 10 103    4 V
R2=5 k
V=6 V
b
The measurement is made with the galvanometer.
60  and 10 k resistors in parallel
are equivalent to an 59.6  resistor.
The total equivalent resistance is
5059.6 , so 1.19x10-3 A of current
flows from the battery.
The voltage drop from a to b is then
measured to be
6-(1.19x10-3)(5000)=0.07 V.
The percent error is.
4 -.07
% Error =
100 = 98%
4
RG=60 
a
G
R1=10 k
R2=5 k
I=1.19 mA
V=6 V
Your opinions? Would you pay for this voltmeter?
b
To reduce the percent error, the device being used as a
voltmeter must have a very large resistance, so a voltmeter
can be made from galvanometer in series with a large
resistance.
a
V
Vab
b

a
RSer
RG
G
b
Vab
Everything inside the blue box is the voltmeter.
Homework hints: “the galvanometer reads 1A full scale” would mean a current of IG=1A would produce
a full-scale deflection of the galvanometer needle.
If you want the voltmeter shown to read 10V full scale, then the selected RSer must result in IG=1A
when Vab=10V.
Example: a voltmeter of resistance 100 k is used to measure
the voltage drop across a 10 k resistor in series with a 6 V
battery and a 5 k resistor (neglect the internal resistance of
the battery). What is the percent error caused by the nonzero
resistance of the voltmeter?
We already calculated the actual
voltage drop (3 slides back).
Vab = IR   0.4 10-3 10 103    4 V
a
R1=10 k
R2=5 k
V=6 V
b
The measurement is now made with the voltmeter.
100 k and 10 k resistors in
parallel are equivalent to an 9090 
resistor. The total equivalent
resistance is 14090 , so 4.26x10-4
A of current flows from the battery.
The voltage drop from a to b is then
measured to be
6-(4.26x10-4)(5000)=3.9 V.
The percent error is.
4 - 3.9
% Error =
100 = 2.5%
4
RV=100 k
a
V
R1=10 k
R2=5 k
I=.426 mA
V=6 V
Not great, but much better. Larger Rser is needed for high accuracy.
b
Measuring Instruments: Ohmmeter
An ohmmeter measures resistance. An ohmmeter is made
from a galvanometer, a series resistance, and a battery.
RG
RSer
G
Everything inside the blue
box is the ohmmeter.
The ohmmeter is connected in parallel with the unknown
resistance with external power off. The ohmmeter battery
causes current to flow, and Ohm’s law is used to determine
the unknown resistance.

R=?
To measure a really small resistance, an ohmmeter won’t
work.
Solution: four-point probe.
A
V
Measure current and voltage separately, apply Ohm’s law.
reference: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/movcoil.html#c4