Ohm’s Law
Objectives:
1_measure the current_voltage curve for a resistor
2_construct a graph of the data from objective 1
3_given a graph of current_voltage for a resistor, determine the resistance
Equipment:
Three resistance R1 = 1.0 kΩ, R2 = 1.5 kΩ, R3 = 2.2 kΩ
One dc ammeter 0_10 mA, wires
Theory:
Potential difference is the work required to transfer electrical charge from one
point to another, and measured in volt. The flow of electrical charge in a circuit is
called current; current is measured in units of amperes.
the relationship between current and voltage is an important characteristic that
defines various electronic devices. the relationship is frequently shown with a
graph , usually the voltage is controlled (the independent variable ) , and the
current is observed (the dependent variable) . Fixed resistors have a straight line
or linear current _voltage curve. This linear relationship illustrates the basic
relationship of ohm’s law. If passed in the current conductor, it generates a
potential difference in this connector is directly proportional to the current
flowing in it is called ohm’s law .Ohm’s law is the most important law of
electronics. it is written in equation form as :
I = V/R
Where I: current, V : voltage , R : resistance
Procedure:
1_ Measure three resistor with listed values of 1.0kΩ, 1.5kΩ , 2.2kΩ . The
measured values is
Component
R1
R2
R3
Listed value
1.0 kΩ
1.5 kΩ
2.2 kΩ
Measured value
1.0 kΩ
1.5 kΩ
2.2 kΩ
Table 5.1
2_ Connect the circuit shown in figure 5_1 (a). The ammeter is in series with the
resistor and forms a single loop as shown in the protoboard wiring diagram in
figure 5_1 (b). The voltmeter is in parallel with the resistor.
3_Connect the power supply for a voltage of 2.0 v. Read the current that is
flowing through the resistor and record it in table 5_2.
4_Connect the power supply for 4.0 v and measure the current. Record the
current in table 5_2. Continue taking current readings for each of the voltage
listed in table 5_2.
Vs =
I=
2.0 v
2.04 mA
3.0 v
4.04 mA
4.0 v
6.04 mA
8.0 v
8.04 mA
10 v
10.02 mA
Table 5.2 (R1)
5_ Replace R1 with R2 and repeat step 3 and 4. Record the data in table 5_3.
Vs =
I=
2.0 v
1.37 mA
3.0 v
2.7 mA
4.0 v
4.04 mA
8.0 v
5.37 mA
10 v
6.7 mA
Table 5.3 (R2)
6_ Replace R2 with R3 and repeat step 3 and 4. Record the data in table 5_4.
Vs =
I=
2.0 v
0.93 mA
3.0 v
1.84 mA
4.0 v
2.75 mA
8.0 v
3.66 mA
10 v
4.56 mA
Table 5.4 (R3)
7_ graph all three I-V curves using the data from tables 5_2, 5_3, and 5_4. The
dependent variable (current) on the y-axis and the independent variable (voltage)
on the x-axis.
I-V Curves for R1
12
10
Current (mA)
8
6
Series1
4
2
0
0
2
4
6
Voltage (v)
8
10
12
I-V Curves for R2
8
7
Current (mA)
6
5
Series1
4
3
2
1
0
0
2
4
6
Voltage (v)
8
10
12
I-V Curves for R3
5
4.5
Current (mA)
4
3.5
3
Series1
2.5
2
1.5
1
0.5
0
0
2
4
6
Voltage (v)
8
10
12
Conclusion:
 The relationship between voltage and current is that the voltage is directly
proportional to the current.
 The relationship between the resistance and the current is that the
resistance is inversely proportional to the Current.
 The relationship between the resistance and the voltage is that the
resistance is directly proportional to the current.
Evaluation and Review Questions:
1_ the slope of a line is the change in the y direction divided by the change in the
x direction. The definition for slope is illustrated in figure 5_2. Find the slope for
each resistor on graph.
For R1 = 1.0kΩ
Slope = change in Y/change in x = (4.04 - 2.04) / (4 - 2) = 1
For R2 =1.5kΩ
Slope = (2.7- 1.37) / (4 -2) = 0.665
For R3 = 2.2kΩ
Slope = (1.84 - 0.93) / (4 - 2) = 0.455
2_ what happens to the slope of the I-V curve for large resistors?
The value of the slope for large resistors equals 0.455 this is small value, and this
shows the inverse proportionality between the resistance and the current.
3_ (a) If the resistance is halved and the voltage is not changed, what will happen
to the current in a resistive circuit?
I = V / 0.5 R
I = 2V / R
Current is increasing because it is inversely proportional to the resistance.
(b) If the voltage is doubled and the resistance is not changed, what will happen
to the current in a resistive circuit?
I = 2V / R
Current is increasing because it is directly proportional to the voltage.
4_ if the current in a resistive circuit is 24 mA and the applied voltage is 48 v, what
is the resistance?
R=V/I
R = 48 / 24 = 2kΩ
5_ what current will flow through a 10Ω resistor with a 5.0 v applied?
I=V/R
I = 5/10 = 0.5 A
Power in Dc Circuits
Objectives:
1_Determine the power in a variable resistor at various settings of resistance
2_plot data for power as a function of resistance. From the plot, determine when
maximum power is delivered to the variable resistor
Equipment:
One 2.2 kΩ resistor
One 10 kΩ resistor
Theory:
When current flows through a resistor, electrical energy is converted into heat.
Heat is then radiated from the resistor. The rate that heat is dissipated is called
power. Power is measured in units of joules per second (j/s) , which defines the
unit called the watt (W) . The power dissipated by a resistor is given by the power
law equation:
P=V*I
By applying ohm’s law to the power, we get two equations for power. These are:
P = I² * R and
P= V² / R
Procedure:
1_ Measure the resistance of R1 .the color – code value is 2.2kΩ. R1 = 2.2kΩ.
2_ Construct the circuit shown in figure 6_1. R2 is a 10 kΩ potentiometer. Connect
the center (variable) terminal to one of the outside terminals. Connect the
potentiometer for 0.5 kΩ. Always remove power when measuring resistance and
make certain you are measuring only the potentiometer resistance.
3_Measure the voltage across R1 and the voltage across R2. Enter the measured
voltages in table 6_1. As a check, make sure that the sum of V1 and V2 is equal to
12 v. then compute the power in R2 using the equations:
P2 = V22 / R2
4_ disconnect the power supply and set R2 to the next value shown in table 6_1.
Reconnect the power supply and repeat the measurements made in step 2.
Continue in this manner for each of the resistance settings in table 6_1. Table 6_1
Variable resistance
setting R2
0.5 kΩ
1.0 kΩ
2.0 kΩ
2.2 kΩ
3.0 kΩ
4.0 kΩ
5.0 kΩ
7.5 kΩ
10 kΩ
V1
(Measured)
9.743 v
8.223 v
6.261 v
5.977 v
5.056 v
4.237 v
3.650 v
2.709 v
2.152 v
V2
(Measured)
2.23 v
3.76 v
5.73 v
6.02 v
6.94 v
7.76 v
8.35 v
9.29 v
9.85 v
Power in R2 :
P2 = V22 / R2
9.945 mW
14.137 mW
16.416 mW
16.472 mW
16.054 mW
15.054 mW
13.944 mW
11.507 mW
9.702 mW
5_ graph the relationship of the power as a function of resistance R2. Since
resistance is the independent variable. Plot it long the x-axis and plot power along
the y-axis. An implied data dissipated R2 without resistance.
The Relationship b/w power and
resistance
18
16
Power(mW)
14
12
10
Series1
8
6
4
2
0
0
5
10
resistance (kΩ)
15
Conclusion:
 Power: Is the time rate of flow of electrical energy in an electrical circuit,
and the unit is measured in watts.
 P = V2 / R, If the voltage is constant inverse relationship between power and
resistance.
 P = I2 R, But if the current is constant in the relationship, then power is
directly proportional to the resistance.
Evaluation and Review Questions:
1_ observe the graph of resistance versus power for your experiment.
Compare the resistance R1 and R2 when power in R2 is a maximum.
The power maximum equals 16.47mW, when R2 = R1
2_ what was happening to the total current in the circuit as R2 was
increasing?
The total current decreases.
3_what was happening to the power in R as the resistance of R2 was
increasing? Explain your answer
1
Variable resistance
setting R2
0.5 kΩ
1.0 kΩ
2.0 kΩ
2.2 kΩ
3.0 kΩ
4.0 kΩ
5.0 kΩ
7.5 kΩ
10 kΩ
V1
(Measured)
9.743 v
8.223 v
6.261 v
5.977 v
5.056 v
4.237 v
3.650 v
2.709 v
2.152 v
P1 = V12 /R1
43.148 mW
30.735 mW
17.818 mW
16.238 mW
11.619 mW
8.160 mW
6.055 mW
3.335 mw
2.105 mW
The power in R1 drop when the resistance of R2 was increasing
4_ A 1.5 kΩ resistor is found to have 22.5 V across it.
(a)What is the current in the resistor?
I=V/R
I = 22.5 / 1.5 =15 mA
(b) What is the power dissipated in the resistor?
P = V* I
P =22.5 * 15 =337.5 mW
5_ what physical characteristic the power rating of a resistor?
1_P = V2 / R, If the voltage is constant inverse relationship between power
and resistance.
2_P = I2 R, But if the current is constant in the relationship, then power is
directly proportional to the resistance.
For further Investigation:
R1 = 2.2 kΩ
V1
(Measured)
9.743 v
8.223 v
6.261 v
5.977 v
5.056 v
4.237 v
3.650 v
2.709 v
2.152 v
I 1 = V1 / R 1
4.428 mA
3.737 mA
2.845 mA
2.716 mA
2.298 mA
1.925 mA
1.659 mA
1.231 mA
0.978 mA
5
4.5
Current 1 (mA)
4
3.5
3
2.5
Series1
2
1.5
1
0.5
0
0
1
2
3
R2 (kΩ)
4
5
12
10
V2 (v)
8
6
Series1
4
2
0
0
2
4
6
R2 (kΩ)
8
10
12