
Laboratory 4: Instrument Characteristics
201010
h) Set the resistance box to 500 Ω.
i) Turn the power supply’s “OUTPUT” ON, without disturbing its voltage setting.
j) Record the value of current indicated on the ammeter.
k) Turn the power supply’s “OUTPUT” OFF.
l) Set the resistance box to 75 Ω.
m) Turn the power supply’s “OUTPUT” ON.
n) Record the value of current indicated on the ammeter.
o) Turn the power supply’s “OUTPUT” OFF.
Analysis: Assuming that the internal resistance of the meter is zero, calculate the theoretical
values for the three currents and compare these to your measured values. Explain any dis-
crepancies and summarize your understanding of the influence of the ammeter on the meas-
urements.
Voltmeter
1. Determining the internal resistance of the digital voltmeter.
Background
To minimize their effect on voltage measurements, voltmeters are designed to have very high
input resistance. This is especially true of the newer digital meters. Historically, one common
method of determining the internal resistance of a volt meter was to connect it in a simple se-
ries circuit with a variable resistor. First, with the variable resistor R
S
set to 0 Ω, measure and
record V
0
, the initial voltage drop across the meter. Then, without adjusting the power supply
or the scale setting on the voltmeter, increase R
S
until the voltmeter indicates ½ V
0
. At that
point half of the applied voltage is dropped across R
S
and half is dropped across the voltme-
ter. Assuming that the voltmeter behaves like a purely resistive load, the resistance of the
voltmeter must be equal to the resistance of R
S
at that setting. That is,
R
Meter
= R
S
when the V
meter
= ½ V
0
Modern digital voltmeters have very high resistances, often higher than the available variable
resistors or resistance boxes. Thus, we must modify the above procedure accordingly.
Assume that the voltmeter in the circuit below behaves electrically like a resistor with resis-
tance R
M
. Define the following variables:
R
S
= the resistance of the variable resistor,
V
0
= the output voltage of the power supply,
V
R
= the voltage drop across the variable resistor R
S
V
M
= the voltage drop across the voltmeter, and
I = the current flowing through the circuit.
Then, by Ohm’s Law and Kirchhoff’s Laws, we must have
V = V
R
+ V
M
where
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