Lab Date
Lab 1: DC Circuits
Student Name1, name1@ufl.edu
Student Name2, name2@ufl.edu
I. Introduction
The purpose of this lab is to allow the students to become comfortable with the
use of lab equipment, through exercises involving simple DC circuits. This lab
introduces us to the use of a power supply, multimeter, and voltmeter. We examine the
properties of DC circuits containing resistors and diodes. We also become familiar with
the use of an oscilliscope and a function generator.
In this lab we will be connecting a DC power supply to circuits containing various
resistive sources (resistors, light bulbs, diodes). We plot voltage (x-axis) vs current (yaxis) and compare the plots for the various setups. This will allow us to understand the
following: how voltage and current are affected by devices that follow Ohm’s law,
devices that are not ideal resistors, and devices that do not follow Ohm’s law. We also
try to calculate and build a Thevenin equivalent circuit.
II. Results
1-1. Ohm’s Law
A. Experimental Setup
We connect a power supply, voltmeter, multimeter, and resistor to the breadboard.
Circuit Diagram:
B. Results
18 k 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
12 k 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
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 slope of each is very
nearly the resistance of the corresponding resistor used. No surprises here!
We used our power supply as the voltmeter and the DMM as the ammeter. Thus, the
voltmeter is not measuring the exact voltage across the resistor because the ammeter is in
the way. One can rearrange the circuit so that the voltmeter is directly across the resistor
(see below), but then the ammeter will also be measuring the current (however small)
through the voltmeter.
The ideal set up would have a voltmeter with infinite resistance and an ammeter with
zero resistance. Then, either set up would work perfectly and both would yield the same
result.
Given a 20k resistor, the voltmeter measured a slightly higher (0.001 V) voltage than the
power supply. With a 20M resistor, this would have been a much greater error. Thus,
the first circuit is better for large resistances and the second is better for small resistances.
1-2. An incandescent lamp
A. Experimental Setup
We replace the resistor from part 1-1 with a lamp.
Circuit Diagram:
B. Results
#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
V vs. I
Current (mA)
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
A. Experimental Setup
We connect two resistors in series to a multimeter. The multimeter is also connected to a
diode, and to a voltmeter.
Circuit Diagram:
B. Results
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, no
current flowed across it, no matter what resistor value we used.
1-4. Voltage Divider
A. Experimental Setup
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:
7.59 V 
Loaded circuit output voltage:
5.02 V 
Short circuit current:
1.474 mA 

The Thevenin equivalent circuit behaves identically to the original circuit, as expected.