# DC Circuits – Ohm's and Kirchhoff's Laws Lab - Rose-Hulman

Ηλεκτρονική - Συσκευές

7 Οκτ 2013 (πριν από 5 χρόνια και 6 μήνες)

593 εμφανίσεις

Scott Jacks Box 1969

Kate Vawter Box 371

DC Circuits

Ohm’s and Kirchhoff’s Laws Lab

Purpose
:

The purpose of this lab is to demonstrate Ohm’s and Kirchhoff’s Laws by
creating circuits and measuring their voltage and calculating their currents.

Equipme
nt
:

DC Circuit Patchboard

Regulated variable DC power supply

Digital Multimeters

Procedure
:

Part A:

1.

Plug in two wires into circuit board into point A and point B.
Using the voltmeter, measure and record the output
voltage.

2.

Measure and record the voltage from A to C.

3.

Measure and record the voltage from C to D.

4.

Measure and record the voltage from A to D.

Part B:

1.

Measure and record potential drop from A to C, and from C to D.

2.

Calculate V
AC

using R= 100 ohms and
I= 6.55 V

Part C:

1.

Set up the circuit

2.

Measure the voltage of dry cell

Calculations:

Circuit 1

Points

Voltage Measured (Volts)

A to B

6.55

A to C

1.31

C to D

5.25

A to D

6.54

To calculate the current of the system, we use Ohm’s law, V=I*R, the
known resistance
from A to C and the measured change in voltage as follows:

V=I*R =>

I=V/R

I=1.31/100 = 0.0131 A

Using the now known current and Ohm’s law, we can compare the known resistance
from C to D with a measured value:

V=I*R =>

R=V/
I

R=5.25/0.131 = 400.8

Now apply a percent difference between the known and measured resistance from C to D
to check the accuracy of our measurements.

( (1.31+5.25)
-
6.54 ) / (1.31+5.25) *100 = 0.3% difference

Also, since these two resistors are c
onnected in series, the voltage drop across the
combination is the sum of the voltage drop across each. According to Kirchoff’s Law,
the sum of the voltage for the system will equal zero.

V
AC

+ V
CD

+ V
BA

= 0

1.31 + 5.25

6.54 = 0.02

Circuit 2

Points

Voltage Measured (Volts)

A to B

6.55

A to C

2.81

C to D

3.75

To calculate the current, use same procedure as circuit 1. Use the measured voltage
difference and known resistance from A to C.

V=I*R =>

I=V/R

I= 2.18/100 = .0281 A

Since the re
sisters are in parallel, each resistor experiences the same voltage but different
current. The two resistors share the current in proportion to their resistance value so
combine to form an equivalent resistance value. This value can be calculated using
O
hm’s Law.

V=I*R =>

R=V/I

R= 3.75/0.0281 = 133.5

A theoretical resistance can be found using parallel resistor properties.

1/R
effective

= 1/R
1

+ 1/R
2

....

1/R
effective

= 1/400 + 1/200 = 3/400

R
effective

= 133.3

Now we can apply a percent dif
ference formula to check the accuracy of the parallel
formula.

( 133.45

133.33 ) / 133.45 *100 = 0.01% difference

Circuit 3

Points

Voltage Measured (Volts)

Current ( Amps )

(E1) A to B

6.55

Not calculated

(E2) G to H

1.53

Not calculated

(R1) A t
o C

2.28

0.0228

(R2) G to E

0.30

0.003

(R3) C to D

4.27

0.01068

(R4) C to E

2.45

0.01225

(R5) E to F

1.83

0.00915

Using the known resistor values and measured voltage differences we can apply Ohm’s
Law to find the current at each resistor. The resul
ts are shown in the table above.

According to Kirchhoff’s first law, the node theorem, the net current flowing at any node
in the circuit must be zero. Therefore, these equation can be should be true:

I
4

(I
2

+ I
5
) = 0

I
1

(I
4

+ I
3
) = 0

I
3

I
5

= 0

0.0
1225

(0.003+0.00915) =
-
0.0001

0.0228

(0.01225+0.01068) = 0.00013

0.01068

0.00915 = 0.00153

According to Kirchhoff’s second law, the loop theorem, the sum of all the voltages in a
closed path must equal zero. We can demonstrate this in each loop by

the following
steps:

V
E1

= V
R1

+ V
R3

6.55 V = 2.28 V+ 4.27 V

6.55 V = 6.55 V

Percent difference = 0%

V
E2

= V
R2

+ V
R5

1.53 V=
-
0.30 V+ 1.83 V

1.53 V = 1.53 V

Percent difference = 0%

V
R3

= V
R4

+ V
R5

4.27 V= 2.45 V + 1.83 V

4.27 V = 4.28 V

Percent differe
nce = 0.002%

Conclusions:

Ohm’s and Kirchoff’s Laws both accurately describe the behavior of electrical current,
voltage and resistance. The slight percent error experienced can be attributed to
resistance in the wires connecting the circuit elements a
nd the digital voltometer. Also,
some readings were so borderline that the number would fluctuate back and forth on the
digital output, our interpretation of the values may not have been perfect.