The Circuits Lab introduces basic concepts that engineers use to design and build electric circuits.
In the Circuits Lab, you will:
1.
Construct electric circuits using a breadboard.
2.
Learn how voltage, current and resistance are measured.
3.
Study Ohm’s Law, Power Law, Kirchhoff’s Current Law and Kirchhoff’s Voltage Law.
4.
Calculate and measure the equivalent resistance of electric circuits.
5.
Understand the polarity of certain electrical components.
Electrical circuits are usually laid out on printed circuit boards. The board is made of an electrically-insulating material. Much of the
"wiring" on a printed circuit board is made from insulated metal paths printed on the board, rather than actual wires. That is why they are called
"printed" circuit boards.
Figure 1 shows the front and back of the printed circuit board inside an iPod Touch. Combinations of resistors, capacitors, diodes, integrated circuits are connected by electrical paths on the boards.
The pins on the components are connected together by wire traces on the board.
Figure 1. The iPod Touch circuit board.
3.1 Current
Electrical current flowing through a wire is similar to water flowing through a pipe. The amount of water flowing through a pipe can be measured in gallons per minute. Electrical current is the amount of charge that moves past a location in the wire per unit time (1 ampere = 1coulomb per second).
3.2 Voltage
Voltage is a measure of the electrical force that causes current to flow. It is also called electric potential because it is a difference in potential energy between two points. Just as pressure causes water to flow through a pipe, voltage difference causes electrons to flow through a wire. Voltage is always measured relative to a reference point.
Circuits Lab - Preparation Page 1

Figure 2 shows a digital voltmeter being used to measure the voltage difference (5.23 Volts) across a
1000 Ohm resistor. In this case, the voltage is constant and the current flowing through the resistor is also constant, therefore we call this a direct current (DC) circuit.
Figure 2. Measuring voltage across a resistor.
3.3 Resistance
A resistor is an electronic component that resists flow of current and obeys Ohm's law. Ohm's law states that the voltage V across a resistance R is proportional to the current I passing through it:
Ohm’s Law 𝑉 = 𝐼 ∗ 𝑅 , (1)
Voltage across resistor (Volts) = Current through resistor (Amps)
×
Resistance (Ohms)
The unit of resistance is the ohm (ohm = volt/ampere, which agrees with the equation). Ohms is sometimes abbreviated by the Greek letter omega
, Ω.
Figure 3 shows a 1000 Ohm resistor and its’ circuit schematic to the right. Notice the polarity of the voltage drop across the resistor if the current I is flowing in the direction indicated by the arrow. Resistors themselves have no polarity, so the resistor can be plugged in any direction.
R
Figure 3. A resistor (left) and the symbol for resistance in a circuit schematic (right).
Circuits Lab - Preparation Page 2

3.4 Light Emitting Diode (LED)
An LED is a device that emits light when current flows through it. It can be used as a light source, or as an indicator. What makes it unique from other components is that current can only flow through it in the direction of the arrow. Therefore, an LED is a polarized component and must always be inserted into a circuit board with the correct orientation.
When you build an LED circuit in lab, you will plug the LED into a circuit board with the positive wire
(longer) connected to the negative wire (shorter) as shown in Figure 4.
2
Figure 4. An LED (left) with its schematic symbol (right).
3.5 Electrical Circuit Schematics
A schematic is an engineering drawing of an electrical circuit.
Each symbol on the schematic represents a component in the circuit. Figure 5 shows the symbols for various electrical components.
The lines in the schematic represent the wires or metal paths on the circuit board that connect the components together. Metals like copper and aluminum, which have very low resistance, are used for circuit connections because they are good conductors of electricity. Figure 6 shows wire connections on a typical printed circuit board.
Figure 5. Schematic symbols.
Figure 6. Wire traces on a printed circuit board.
Circuits Lab - Preparation Page 3

4.1 Ohm’s Law The voltage difference V (Volts) across a resistor is equal to the product of the current I
(Amperes) and the resistance R (Ohms):
Ohms Law 𝑉 = 𝐼 ∗ 𝑅 , (2)
4.2 Power Law The power dissipated P (Watts) in a component with resistance R is equal to product of the voltage V (Volts) across the component and the current I (Amperes) flowing through it:
Power Law
Using Ohm’s Law,
Power Law
𝑃 = 𝐼 ∗ 𝑉
,
𝑃 = 𝐼 2 ∗ 𝑅 =
𝑉 2
𝑅
,
(3a)
(3b)
4.3 Calculation of current, voltage and power dissipated in the circuit
Let’s use the Ohm’s Law to solve sample problems.
Consider a resistor schematic (Figure 7) that has a resistor R = 10 kΩ and a voltage input V = 10 volts. Current, I R Voltage, V
Calculate: the current flowing through the resistor.
𝐼 =
10 𝑉𝑜𝑙𝑡𝑠
10 𝑘Ω
=
10 𝑉𝑜𝑙𝑡𝑠
10 , 000 Ω
= 0.001 𝐴𝑚𝑝 = 1 𝑚𝐴
Figure 7. Resistor schematic.
For the case of the same resistor, R = 10 kΩ but with current of 0.5 mA flowing through it,
Calculate: the voltage across the resistor
Current, I
𝑉 = (0.5 𝑚𝐴 ) ∗ (10 𝑘Ω ) = (0.0005 𝐴 ) ∗ (10,000 Ω )
𝑉 = 5 𝑉𝑜𝑙𝑡𝑠 15 Volts
BATTERY
Let’s use Ohm’s Law and Power Law to solve a sample problem.
The circuit in Figure 8 has a battery voltage V = 15 Volts and a resistor R = 5 Ohms.
R = 5 Ohms
Figure 8. Basic Circuit Schematic.
Calculate: (1) Current across the circuit, and (2) Power dissipated in the resistor.
(1) Current, using Ohm’s Law,
(2) Power , using Power Law,
=
𝑉
𝑅
=
15 𝑉𝑜𝑙𝑡𝑠
5 𝑂ℎ𝑚𝑠
= 3
𝑃 = 𝑉 ∗ 𝑖 = 15 𝑉 ∗ 3 𝐴 = 45 𝑊𝑎𝑡𝑡𝑠
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4.4 Kirchhoff’s Voltage Law:
In the circuit of Figure 9 below, the battery makes electrons flow through the wires and each resistor in a loop. Electrons carry negative charge and leave the negative terminal of the battery, returning to the positive terminal. However, conventional current is defined as the movement of positive charge, and so we define the positive flow of current as the direction shown of the arrow for the conventional current I .
Since the same current I flows through each of the resistors, it creates the voltage differences across each resistor, shown by the three voltages V
1
, V
2
and V
3
.
Current, I
Vbat
R1
R2
V1
V2
R3
V3
Figure 9. Kirchhoff's voltage law sample circuit.
Therefore, the battery voltage V bat is equal to the sum of the voltage difference across the resistors, or
Kirchhoff’s Voltage Law 𝑉 𝑏𝑎𝑡
= 𝑉
1
+ 𝑉
2
+ 𝑉
3
, (4)
4.5 Kirchhoff’s Current Law Since voltage is the same across every resistor connected in parallel, the sum of currents entering into a node is equal to the sum of currents leaving a node (a node in a circuit is the junction of 3 wires shown usually as a dot), or
Kirchhoff's Current Law 𝐼
0
= 𝐼
1
+ 𝐼
2 , (5)
I2 I0
I1
Vbat R1 R2
Circuits Lab - Preparation
Figure 10. Kirchhoff's current law sample circuit.
Page 5

4.6 Equivalent Resistance
For calculation purposes, several resistors can be replaced by a single “equivalent” resistor ( R eq
).
4.6.1 Resistors Connected in series
Since the current is the same through all the resistors connected in series, then the equivalence resistance is the sum of all the resistors, or
Equivalent Resistance (series) R eq
(Series) = R
1
+ R
2
+ R
3 (6a)
Current, I
R1
V1
Current, I
Vbat
R2
V2 Vbat Req (Series)
R3
V3
Figure 11. Equivalent resistance of resistors in series.
4.6.2 Resistors Connected in Parallel
Since voltage is the same across every resistor connected in parallel, the total current is the sum of the currents flowing through the individual resistors.
Vbat
I0
R1 R2 R3 Vbat
I1 I2
I3
Figure 12. Equivalent resistance of resistors in parallel.
I0
Req (Parallel)
Circuits Lab - Preparation Page 6

For resistors in parallel,
Equivalent Resistance
(parallel)
1
R eq
=
1
R
1
+
1
R
2
+
1
R
3
Note that the equivalent resistance is 1/R eq
, so in order to find R eq
, the inverse must be taken.
(6b)
4.6.3 Calculation of current, power dissipated and equivalent resistance
Let us consider an electric circuit schematic (shown in Figure 13) that has a battery voltage of V
BAT
=15 volts connected to resistors R
1
= 100 ohms and R
2
= 200 ohms in parallel.
Calculate: (1) Current for resistor R
1
, (2) Power dissipated P
R1 for resistor R
1
, and (3) Equivalent resistance R eq of the overall circuit.
I0 I2
Vbat
15 Volts
R1
100 Ohms
I1
R2
200 Ohms
Figure 13. Basic circuit with resistors connected in parallel.
(1) Current across resistor R
1 .
Keep in mind that in a parallel circuit the battery voltage across R
1
and
R
2 is the same V
BAT
=15volts. Therefore, using Ohm’s Law for resistor R
1
𝐼
1
=
𝑉
𝑅
1
=
15
100
= 0.15 𝐴𝑚𝑝
= 150 mA
(2) Power across resistor R
1
. using Power Law,
𝑃
𝑅
1
= 𝑉 ∗ 𝐼 = 15 ∗ 0.15 = 2.25 𝑊𝑎𝑡𝑡𝑠
(3) Equivalent Resistance Req. Substituting R
1
and R
2
into Equation 6b, the equivalent resistance Req is calculated,
𝐞𝐪
1
=
R
1
+
1
R
2
=
1
100 Ω
+
1
200 Ω
=
.
𝑅 𝑒𝑞
=
1
0.015
= 66.7
Circuits Lab - Preparation Page 7

A special approach must be taken when adding resistors that are in both parallel and series within the same circuit. Determining the equivalent resistance requires converting either various parallel resistances into series or vice versa. In other words, the goal is to have all resistors in parallel or series after finding additional equivalent resistances.
Let’s see the example shown in Figure 14. We have a battery voltage connected to 3 resistors R
1
, R
2
and
R
3
. Resistors R
2
and R
3 are connected in series and R
1 is connected parallel to resistors R
2
and R
3
.
I0 I2
R2
Vbat R1 I1
R3
Figure 14: A series and parallel resistances circuit
Vbat
I0
R1
I1
I2
R23
Figure 15: A series and parallel resistances circuit converted to only a parallel resistances circuit
I0
Vbat Req
Figure 16: Simplified circuit with equivalent resistance
In order to get the overall equivalence resistance, we need to simplify the circuit step by step. The first step will be to combine R
2
and R
3 to create Figure 15. In order to obtain R
23
, we will use the equivalence resistance equation 6a for resistors R
2
and R
3 connected in series. Notice that we have R
1
and R
23
in parallel; therefore, we can use equation 6b to obtain the equivalent resistance R eq
and simplify the circuit as shown in Figure 16.
The following equation 6c shows the progression of two series resistances are converted to parallel for ease of calculation,
Equivalent Resistance
1
𝑅 𝑒𝑞
=
1
𝑅
1
+
𝑅
2
1
+ 𝑅
3
=
1
𝑅
1
+
1
𝑅
2 , 3
(6c)
Circuits Lab - Preparation Page 8

5.1 Bread Board
Bread boards are commonly used with circuits for prototyping because of their ease of use and reusability. Wires from the components are placed into contact with metal strips in order to connect them in series. As you can see from Figures 17 and 18 below, holes are connected vertically and horizontally on the board, which allows connections to be made easily. Note that there are gaps in between the metal pieces, so the strips are not touching. Therefore, at times it may be necessary to jump the gap with a wire.
At the top, there are power and ground “busses”. These are useful for connecting banana leads from multimeters or power sources.
Figure 17: The front of a breadboard showing the connections between the different alignments of holes
5.2 Digital Multimeter (DMM)
Figure 18: The back of the breadboard showing the metal connections correlating to the way they are on the front
Digital multimeters are designed to measure electrical properties. They are most commonly used for measuring volts, current, and resistance. However, today the DMM can measure many others including capacitance, inductance, and transistors. Typically, the leads on the multimeter are red and black with red being the positive and black being the negative. There are several settings on the multimeter that allow it to be so versatile. The DCV setting stands for direct current voltage, which will be used to measure the difference in voltage between two points. Recall that t he Ω symbol stands for Ohms, the units for resistance. The other DMM settings can be used, but these ones previously listed will be used in conjunction with the lab. Note that current can always be calculated using Ohm’s law from measured resistance and voltage values. Never measure CURRENT using the DMM. The current should always be calculated from measured voltage and resistance values.
Circuits Lab – Pre-Lab Assignment Page 9

DC Voltage
(Volts)
Resistance
(Ohms)
Figure 19: Digital Multimeter (DMM) with probes attached
In order to finish your preparation for Circuits Lab, watch the video on how to use the DMM and take the
Circuits Lab quiz on Carmen.
Circuits Lab – Pre-Lab Assignment Page 10

15 Volts
Vbat
This is an individual assignment.
Solve the five problems below and hand it in at the beginning of the Circuits Lab.
Problem 1. Ohm’s Law. For the circuit below, calculate the value of the resistor R which would cause the current of 2.5 mA to flow in the circuit. Show your calculations. (4 points)
I
2.5 mA
R V res
R = _________________________
What voltage would you measure across the resistor? V res
= ________________________
5 Volts
Vbat
Problem 2. Kirchhoff’s Voltage Law.
For the circuit below, calculate: (1) the equivalent resistance,
(2) the current I flowing in the circuit, (3) the voltages V
1
and V
2
, and (4) verify that V
Show all calculations. (Note: 1 kΩ = 10 3 Ω). (5 points)
1
+ V
2
= V bat
.
I
R1
10k
R2
20k
V1
V2
R eq
= _____________________
I = _______________________
V
1
= _____________________
V
2
= _____________________
V
1
+ V
2
= _________________ = V bat
?
Circuits Lab – Pre-Lab Assignment Page 11

Problem 3. Power Law. For the circuit in Problem 2, calculate the power dissipated in each resistor, and the total power generated. Show all calculations. (3 points)
6 Volts
Vbat
P
1
= _______________________ P
2
= _______________________ P
Total
= _______________________
Problem 4. Equivalent Resistance.
Calculate the total equivalent resistance of the circuit below. Also, calculate the total current supplied by the battery, I
0
. Show all calculations. (3 points)
I0
R1
3 Ohms
I1
R2
9 Ohms
I2
R3
18 Ohms
I3
R eq
= ______________________
I
0
= ________________________
Problem 5. Kirchhoff’s Current Law. For the circuit in Problem 4, calculate the currents, I
1
, I
2
, and I
3
.
Does I
1
+ I
2
+ I
3
= I
0
(from Problem 4)? Show all calculations. (3 points)
I
1
= _______________________
I
2
= _______________________
I
3
= _______________________
Does I
1
+ I
2
+ I
3
= _________________ = I
0
?
Circuits Lab – Pre-Lab Assignment Page 12