PURPOSE

We will determine the voltage and current of many simple circuits. We will experiment using a multimeter to find the circuit’s current and voltage and the resistance of a resistor. We will demonstrate the use of Ohm’s Law, Kirchoff’s Voltage Law, and Kirchoff’s Current Law and its application to the circuits we will be experimenting on in this lab.

MATERIALS

two 1 K ohm resistors two digital multimeter (DMM)

power supply (+20 V) wire

two 1 M ohm resistors protoboard

two banana jacks two alligator clips

EXPERIMENT 1:

Procedure

- use a 1K ohm resistor with a color code of brown, black, red, and gold

- gold represents 5% tolerance

- connect one lead of the resistor to one ohmmeter terminal, and the other lead to the other terminal

- put resistor on protoboard or use alligators jacks to connect to leads on the ohmmeter

Analysis

R1 = 1 K ohm

tolerance of resistor is 5%

range of resistor is 950 ohm to 1050 ohm (1 K ohm +- 50 ohm)

Measurements

measured value: R1 = 988 ohm

Comparisons

We find that R1 is within the 5% tolerance value. 988 ohm is indeed in the range between 950 ohm to 1050 ohm. We know this by the gold (5% tolerance) in the resistor’s 4^{th} color band. So the measured value is not the expected value, but indeed within the resistor’s tolerance value.

(problem 2)

Procedure

- use +5V from the power supply using a +20 V scale

- use DC volts on the multimeter, measure voltage across the power supply

- then switch meter leads to the other jack

Analysis

Vpower supply = 5.00 V

Vswitched leads = - 5.00 V

Measurement

Vpower supply = 5.00 V

Vswitched leads = - 5.00 V

Comparisons

The voltage across the power supply and the voltage with the leads switched have the same magnitude, but have a different current direction. The COM port gives a negative direction of current when it is connected to the positive side of the power supply while the AMP jack connects to the negative side.

EXPERIMENT 2

Procedure

- put a 1 K ohm resistor on protoboard and connect protoboard to the power supply with jacks

- set the power supply voltage to +5V (with the +20V scale option on the power supply)

- measure the power supply voltage with the multi-meter

- multimeter setting is to DC volts to measure the voltage

Analysis

V = 5.00 V

I = V/R = 5 V / 1 K ohm = 5 mA (Ohm’s Law)

P = IV = (5 mA) x (5 V) = 25 mW (power dissipation)

- the power dissipation of the circuit should be below the ¼ W resistor power rating

Measurements

the voltage reading on the voltmeter = 5.00 V

voltage reading on the power supply = 5.02 V

- because the voltmeter voltage reading is much more accurate, we will use that in our experiment, calculations, comparisons, and conclusions

I = 5.00 mA

P = IV = (5.00 mA) x (5.00 V) = 25 mW

Circuit diagrams

Comparisons

The power dissipation of the measured valued was the exact value of what we predict. The 25 mW power dissipation is indeed below the ¼ W power rating.

EXPERIMENT 3

Procedure

- add a second 1 K ohm resistor so the two resistor will be in a series circuit

- measure the voltages across each resistor

- find if circuit obeys KVL

- measure the current of the series circuit

- find Req by using I and V1 with V2

- measure the voltage with the leads in opposite leads across R1

Analysis

I = V/(R1 + R2) = 5 V / ( 1 K ohm + 1 K ohm) = 2.5 mA

R1 = 1000 ohm

R2 = 1000 ohm

Req = R1 + R2 = 2 K ohm

Req = V / I = 5 V / 2.5 mA = 2 K ohm

V1 = IR1 = (2.5 mA) x (1 K ohm) = 2.5 V

V2 = IR2 = (2.5 mA) x (1 K ohm) = 2.5 V

KVL: V = V1 + V2

5 V - IR1 - IR2 = 0

5 V - 2.5 V = 2.5 V = 0

KVL applies to the circuit since the sum of the voltages around the loop = 0

V3 = (- 2.5 mA) x (1 K ohm) = - 2.5 V

Measurements

R1 = 988 ohm

R2 = 985 ohm

V1 = 2.499 V

V2 = 2.497 V

Req = R1 + R2 = 988 ohm + 985 ohm = 1973 ohm

I = 2.51 mA

Req = V / I = (2.499 V + 2.497 V) / 2.51 mA = 1990 ohm

Req is within 0.85% between total R and V/I calculations

V3 = - 2.499 V

KVL: 5.00 V - 2.499 V - 2.497 V = 0.004 V

- 0.004 V is rounded off to the value of 0 V

Circuit diagrams

Comparisons

The measured voltages across the resistors were somewhat lower than our predicted values. The equivalent resistances however have a 0.85% difference, primarily caused by the resistance tolerance of 5%. We can say that the 0.85% difference is within the range of series circuit since the resistor does have the tolerance factor. V3 , which has there leads switch to opposite jacks, indeed have a negative value, but has the same value as do V1. The circuit does obey KVL, because the sum of the voltages around the circuit (loop) equals to zero when rounded.

EXPERIMENT 4

Procedure

- remove R2 and place it into parallel position to R1

- measure the currents through each resistor with an ampmeter

- calculate the equivalent resistance of the circuit using the resistors and the with I

- measure the current flowing through the whole circuit

- measure I3 as the jacks are switched at the power supply

Analysis

R1 = 1 K ohm

R2 = 1 K ohm

I1 = V / R1 = 5 V / 1 K ohm = 5 mA

I2 = 5 mA

I = I1 + I2 = 5 mA + 5 mA = 10 mA

Req = (2.5 V / 10 mA) + (2.5 V / 10 mA) = 500 ohm

Req = V / I = 5 V / 10 mA = 500 ohm

KCL: I = I1 + I2

10 mA - 5 mA - 5 mA = 0

- circuit should obey KCL since sum of current around the circuit is zero

I3 = V / R = (- 5 V) / (500 ohm) = - 10 mA

Measurements

R1 = 988 ohm

R2 = 985 ohm

I1 = 5.02 mA

I2 = 5.01 mA

I = 10.03 mA

Req = 1/((1/988 ohm) + (1/985 ohm)) = 493.2 ohm

Req = V / (I1 + I2) = 5 V / 10.03 mA = 498.5 ohms

Req difference is 1.1%

KCL: I = I1 + I2

10.03 mA - 5.02 mA - 5.01 mA = 0

V3 = - 10.03 mA

Circuit diagrams

Comparisons

KCL is obey in this circuit. The total current through the current equals to the current through both resistors. The equivalent resistances has a difference again of about 1.1%. Due to some disaccuracy with the power supply and the tolerance values of the resistors, this difference between the Req’s may occur. A light difference in resistance doesn’t make a difference in the current of the circuit. As for I3, the current is the same as the current flowing through the whole circuit, for the exception that the current is flowing in the opposite direction.

EXPERIMENT 5

Procedure

- use two 1 M ohm resistors in series

- measure the voltages through each resistor

- measure the current of the whole circuit

Analysis

R1 = 1 M ohm

R2 = 1 M ohm

tolerance of resistor is 5%

range of resistor is 950000 ohm to 1050000 ohm (1 M ohm +- 50000ohm

I = V / (R1 + R2) = 5 V / (1 M ohm + 1 M ohm) = 2.5µA

V1 = V2 = IR1 = (2.5µA) x (1 M ohm) = 2.5 V

Measurements

R1 = 968 K ohm

R2 = 976 K ohm

V1 = 2.412 V

V2 = 2.446 V

KVL: 5 V - 2.412 V - 2.446 V = 0.142 = 0, obeys KVL

-cannot measure current value because the ampmeter can’t measure such a small number,

we will solve the current through Ohm’s Law

I = V / (R1 + R2) = 4.858 V / (968 K ohm + 976 K ohm) = 2.499 µA

Comparisons

Because of the very high resistances, we couldn’t measure the value of the current since its value will be too small for the ampmeter reading. Instead, we use Ohm’s Law to solve the circuit’s current. The voltages across the resistors were smaller than predicted, probably because of the resistance tolerance. KVL can still be observed: 5 V - 2.412 V - 2.446 V = 0.142 V, where the final result can be round to zero because of some accuracy problems in the circuit.

In this exercise, we learned that the higher the resistance, the lower the current in the circuit, as long as the voltage through the circuit remains constant.

CONCLUSION

In this lab, we had learned how to measure the voltage through the circuit and it’s resistors. We learn how to put resistors into series and parallel and see how each situation cause a different current to develop. We learn how to use the multimeters and the power supply in the lab. We found that there are many area of experiment error in the equipment, including the tolerance levels in the resistors. We learn the use of Ohm’s Law; determining that the higher the resistance, the current decreases, as long as the voltage stands the same. In series, we learn that resistance adds up to find the equivalent resistance. In parallel, we learn that resistance adds up the inverses and use the inverse of that to find the equivalent resistances. We learned that by switching the two multimeter connecter leads from positive to negative jack and vice versa, the current flows in the oppostie direction, and the value of the voltage will become negative. We learn to build easy circuits on our protoboards and get our hands into electrical engineering.