Lab 3: Circuits Lab Preparation
1. Overview of the Basic Circuits Lab
The Circuits Lab introduces basic concepts that engineers use to design and build electric circuits.
In the Circuits Lab, you will:
1.
2.
3.
4.
5.
Construct electric circuits using a breadboard.
Learn how voltage, current and resistance are measured.
Study Ohm’s Law, Power Law, Kirchhoff’s Current Law and Kirchhoff’s Voltage Law.
Calculate and measure the equivalent resistance of electric circuits.
Understand the polarity of certain electrical components.
2. Circuits in Technology
Electrical circuits are usually laid out on printed circuit boards. The board is made of an electricallyinsulating 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.
3. Measuring Electrical Circuits
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).
Figure 1. The iPod Touch circuit board.
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.
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.
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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
(1)
𝑉 = 𝐼 ∗ 𝑅,
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.
Current, I
R Voltage, V
Figure 3. A resistor (left) and the symbol for resistance in a circuit schematic (right).
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3.4 Light Emitting Diode (LED)
R2
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
1K
D3
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.
1N4148
Q1
D2
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.
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OR
4. Electric Circuit Laws
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):
(2)
𝑉 = 𝐼 ∗ 𝑅,
Ohms Law
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:
(3a)
𝑃 = 𝐼 ∗ 𝑉,
Power Law
Using Ohm’s Law,
𝑃 = 𝐼2 ∗ 𝑅 =
Power Law
𝑉2
,
𝑅
(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
For the case of the same resistor, R = 10 kΩ but with current of 0.5
mA flowing through it,
Figure 7. Resistor schematic.
Calculate: the voltage across the resistor
Current, I
𝑉 = (0.5 𝑚𝐴) ∗ (10 𝑘Ω) = (0.0005 𝐴) ∗ (10,000 Ω)
𝑉 = 5 𝑉𝑜𝑙𝑡𝑠
Let’s use Ohm’s Law and Power Law to solve a sample
problem.
15 Volts
BATTERY
R = 5 Ohms
The circuit in Figure 8 has a battery voltage V = 15 Volts
and a resistor R = 5 Ohms.
Figure 8. Basic Circuit Schematic.
Calculate: (1) Current across the circuit, and (2) Power
dissipated in the resistor.
𝑉
15 𝑉𝑜𝑙𝑡𝑠
(1) Current, using Ohm’s Law,
𝐼 =
(2) Power, using Power Law,
𝑃 = 𝑉 ∗ 𝑖 = 15 𝑉 ∗ 3𝐴 = 45 𝑊𝑎𝑡𝑡𝑠
Circuits Lab - Preparation
𝑅
=
5 𝑂ℎ𝑚𝑠
= 3 𝐴𝑚𝑝𝑠
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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 V1, V2 and V3.
Current, I
V1
R1
Vbat
V2
R2
V3
R3
Therefore, the battery
the sum of the voltage
resistors, or
Figure 9. Kirchhoff's voltage law sample circuit.
𝑉𝑏𝑎𝑡 = 𝑉1 + 𝑉2 + 𝑉3,
Kirchhoff’s Voltage Law
voltage Vbat is equal to
difference across the
(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
(5)
𝐼0 = 𝐼1 + 𝐼2,
Kirchhoff's Current Law
I2
I0
I1
Vbat
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R1
R2
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Figure 10. Kirchhoff's current law sample circuit.
4.6 Equivalent Resistance
For calculation purposes, several resistors can be replaced by a single “equivalent” resistor (Req).
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
R eq (Series) = R1 + R 2 + R 3
Equivalent Resistance (series)
(6a)
Current, I
V1
R1
Vbat
Current, I
V2
Vbat
Req (Series)
R2
V3
R3
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.
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I0
I0
Vbat
R1
R2
I1
Vbat
R3
I2
Req (Parallel)
I3
Figure 12. Equivalent resistance of resistors in parallel.
For resistors in parallel,
1
1
1
1
= + +
R eq R 1 R 2 R 3
Equivalent Resistance
(parallel)
(6b)
Note that the equivalent resistance is 1/Req, so in order to find Req, the inverse must be taken.
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 VBAT=15
volts connected to resistors R1 = 100 ohms and R2 = 200 ohms in parallel.
Calculate: (1) Current for resistor R1, (2) Power dissipated PR1 for resistor R1, and (3) Equivalent
resistance Req of the overall circuit.
I2
I0
Vbat
15 Volts
R1
100 Ohms
I1
R2
200 Ohms
Figure 13. Basic circuit with resistors connected in parallel.
(1) Current
across resistor
R1. Keep in
mind that in a
parallel circuit the battery voltage across R1 and R2 is the same VBAT =15volts. Therefore, using
Ohm’s Law for resistor R1
𝐼1 =
𝑉
𝑅1
=
15
100
= 0.15 𝐴𝑚𝑝 = 150 mA
(2) Power across resistor R1. using Power Law,
𝑃𝑅1 = 𝑉 ∗ 𝐼 = 15 ∗ 0.15 = 2.25 𝑊𝑎𝑡𝑡𝑠
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(3) Equivalent Resistance Req. Substituting R1 and R2 into Equation 6b, the equivalent resistance Req is
calculated,
𝟏
1
1
1
1
= +
=
+
= 𝟎. 𝟎𝟏𝟓
𝐑 𝐞𝐪 R1 R2 100 Ω 200 Ω
𝑅𝑒𝑞 =
1
0.015
= 66.7 Ω
4.7 Resistances in Parallel and Series Together
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 R1, R2 and
R3. Resistors R2 and R3 are connected in series and R1 is connected parallel to resistors R2 and R3.
I2
I0
I2
I0
R2
I1
Vbat
R1
I1
Vbat
R1
R23
R3
Figure 14: A series and parallel resistances
circuit
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 R2 and R3 to create Figure 15. In order to obtain R23, we will use the equivalence
resistance equation 6a for resistors R2 and R3 connected in series. Notice that we have R1 and R23 in
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parallel; therefore, we can use equation 6b to obtain the equivalent resistance Req 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
Circuits Lab - Preparation
1
1
1
1
1
=
+
=
+
𝑅𝑒𝑞 𝑅1 𝑅2 + 𝑅3 𝑅1 𝑅2,3
(6c)
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5. Equipment
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
Figure 18: The back of the breadboard showing the
metal connections correlating to the way they are
on the front
5.2 Digital Multimeter (DMM)
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 the 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.
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DC Voltage
(Volts)
Resistance
(Ohms)
Figure 19: Digital Multimeter (DMM) with probes attached
6. Further preparation assignment
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
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Lab 3: Circuits Lab - Pre-lab Assignment
Name ________________________________ Team ______ Seat No. _______
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
15 Volts
Vbat
R
Vres
R = _________________________
What voltage would you measure across the resistor?
Vres = _____________
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 V1 and V2, and (4) verify that V1 + V2 = Vbat .
Show all calculations. (Note: 1 kΩ = 103 Ω). (5 points)
I
R1
10k
V1
Req = _____________________
5 Volts
Vbat
I = _______________________
R2
20k
V2
V1 = _____________________
V2 = _____________________
V1 + V2 = _________________ = Vbat ?
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Name ________________________________ Team ______ Seat No. _______
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)
P1 = _______________________ P2 = _______________________ PTotal = _______________________
Problem 4. Equivalent Resistance. Calculate the total equivalent resistance of the circuit below. Also,
calculate the total current supplied by the battery, I0. Show all calculations. (3 points)
I0
6 Volts
Vbat
R1
3 Ohms
I1
R2
9 Ohms
I2
R3
18 Ohms
I3
Req = ______________________
I0 = ________________________
Problem 5. Kirchhoff’s Current Law. For the circuit in Problem 4, calculate the currents, I1, I2, and I3.
Does I1 + I2 + I3 = I0 (from Problem 4)? Show all calculations. (3 points)
I1 = _______________________
I2 = _______________________
I3 = _______________________
Does I1 + I2 + I3 = _________________ = I0 ?
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