Lab #1: DC Circuits
1-1. Ohm’s Law
Circuit Diagram:
12 kΩ Resistor
V (V)
I (mA)
1.00
0.084
2.00
0.169
2.99
0.252
4.00
0.338
4.99
0.422
10.00
0.846
18 kΩ Resistor
V (V)
I (mA)
1.00
0.055
2.00
0.112
3.00
0.168
4.00
0.224
5.00
0.280
10.00
0.561
V vs. I
Current (mA)
1
0.8
0.6
18k
12k
0.4
0.2
0
0
2
4
6
Voltage (V)
8
10
12
The curves are clearly linear, as expected by Ohm’s Law. The slopes are 17.82 ohm and 11.82
ohm, close to the resistance of the corresponding resistor used.
In the circuit diagram above we can see that the voltmeter is measuring any voltage drop across
the resistor or ammeter. If the ammeter were ideal, then its internal resistance would be zero and
consequently there would be no voltage drop across it. In reality, the ammeter will have nonzero
resistance. We can try to fix the problem by moving the terminals of the voltmeter to cross only
the resistor. However, this introduces a new problem. The voltmeter is now directly in parallel to
the resistor, so that the measured current will now be the current passing through the resistor plus
the current passing through the voltmeter. If the voltmeter were ideal, it would have infinite
resistance and there would be no current passing through it. In that case, our measurements would
not be affected.
It is possible to measure the internal resistance of the ammeter and voltmeter (configured for this
experiment) by placing each device in series with a resistor comparable to the internal resistance
of the device. In the case of the voltmeter, we placed it in series with a 20M resistor and set the
input voltage at 5.00 V. The voltage measured by the voltmeter was only 1.66 V. We can
compute the internal resistance knowing the voltage divider equation:
Vmeas = [ Rint / (Rint + Rext ) ] Vin
Solving, Rint = [ Vmeas / (Vin - Vmeas) ] Rext = [1.66/3.34]20 Mohm = 9.94 Mohm ~ 10
Mohm.
In the case of the ammeter, we placed it in series with a 1kohm resistor and set the input voltage at
0.20 V. We know Imeas (Rext + Rint) = Vin, or, after solving, Rin = Vin/Imeas-Rext = (0.20
V)/(.173 mA) -(1.0 k ohm)~ 150 ohm
Knowledge of the internal resistance of our measuring devices allows us to estimate the error
for the two possible positions our voltmeter may be in. If the voltmeter is across only the
resistor, then it draws about Rext /(Rext + Rint) = (20 k)/(10 M) = 0.002 (or 0.2%) of the current
away from the 20k resistor. On the other hand, if the voltmeter is across both resistor and
ammeter, then the ammeter draws about Rint /(Rext + Rint) =150/20k = 0.0075 (or 0.8%) of
the voltage away from the resistor. Thus, in our case, it is preferable by a small margin to place
the voltmeter across the resistor. This conclusion would be reversed if we were trying to
measure a 20M resistor since then placing the voltmeter directly across the resistor will have it
draw out 1/3 of the current.
1-2. An incandescent lamp
Circuit Diagram:
#47 Lamp
V (V)
I (mA)
1.00
52.6
2.00
77.8
3.00
98.3
4.00
115.5
5.00
132.0
Current (mA)
V vs. I
140
120
100
80
60
40
20
0
0
1
2
3
Voltage (V)
4
5
6
The curve is not linear. Instead, it curves down slightly towards the voltage axis. Thus, the light
bulb does not obey Ohm’s Law. Since resistance can be thought of as the derivative of voltage
with respect to current, the light bulb has no clear, constant resistance. Instead, one can only
specify the resistance as a function of the current. One possible explanation for the funny shape
of the curve is that as voltage increases, the lamp filament heats up. This heating can directly
change the resistance of the filament. Another possible explanation is that as the light bulb
grows brighter and heats up, more power is extracted from the circuit. This power is tapped
from the current, causing the curve to slope down as it does.
1-3. The Diode
Circuit Diagram:
1N914 Diode
V (V)
I (mA)
0.37
0.009
0.45
0.045
0.48
0.088
0.53
0.210
0.54
0.297
0.56
0.403
V vs. I
Current (mA)
0.5
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
Voltage (V)
0.8
1
0.8
1
V vs. I Semi-Log Plot
Current (mA)
1
0.1
0.01
0.001
0
0.2
0.4
0.6
Voltage (V)
The linear plot curve looks exponential. The semi-log plot curve looks linear. Thus, we can
safely assume the V-I dependence is exponential. If you were to put 5 volts across the diode, as
the curve tells you, the current would increase to a gigantic amount. This would definitely blow
out the diode. When we reversed the direction of the diode, almost zero current (measured to be
< ___A) flowed across it, no matter what resistor value we used.
1-4. Voltage Divider
Circuit Diagram:
Open
Open circuit output voltage:
Loaded circuit output voltage:
Short circuit current:
Loaded
7.56 V
5.05 V
1.497 mA
VTh = Vopen = 7.56 V
RTh = VTh / Ishort = 7.56V / 1.497mA = 5.05k
Thevenin Circuit Diagram:
Open circuit output voltage:
Loaded circuit output voltage:
Short circuit current:
7.59 V √
5.02 V √
1.474 mA √
The Thevenin equivalent circuit behaves identically to the original circuit, as expected.