ELT-2071 Series Circuits, Kirchoff Voltage Law, Voltage Dividers Isaac Parenteau
1/19/2007
Objective:
1. To study the relationship between voltage and resistance in series circuits.
2. To study how Kirchoff’s Voltage Law applies to series circuits.
3. To study how series circuits can be analyzed with a voltage divider
Equipment:
12V DC Power Supply
Digital Multimeter
Decade Resistance Box (Voltage Divider)
Various Components
3.17V DC Battery Pack
Proto board
Part 1: Series Circuit
Theory:
Electrical parameters in a series circuit behave in the following manner:
1. Voltages drop across each resistor.
2. Current is the same throughout the circuit
3. Total resistance is the sum of all resistors
The voltage drop across each resistor adds up to the applied voltage.
Wiring schematic:
Figure 1:
Figure 2:
Data Table:
Circuit in Figure 1
Voltage
Measurement
12.01V
E
2.45V
ER1
9.56V
ER2
Circuit in Figure 2
Voltage
Measurement
12.01V
E
1.23V
ER1
4.84V
ER2
5.94V
ER3
10.78V
Ebd
6.08V
Eac
For the first part of part one (Figure 1), I measured the voltage of the whole system
which was 12.01 Volts. Then, I used the Digital Multimeter and set it on the volt setting
to measure the voltage drop across each resistor. For R1, I got a voltage drop of 2.45V
and for R2 I got a voltage drop of 9.56V. When the two voltage drops are added together
I get a combined total of 12.01V which is the same as the input voltage.
For the second part of part 1 (Figure 2), I added another resistor of 470 ohms to the
circuit, which increased the total resistance of the circuit and decreased the voltage drop
across the previous two circuits. I measured the voltage drop across R1, R2, and R3, and
got 1.23V, 4.84V, and 5.94V respectively. Then I measured the voltage across R2 and R3
(subscripted Ebd) and got 10.78V. I also did the same thing across R1 and R2 (subscripted
Eac) and that gave me a measurement of 6.08V.
I also observed that Eac = Ebd + Eab -Ecd because in a series circuit the total resistance is
the sum of all the resistors, so in the case we can also use the Resistors to see how this
works. It also allows me to use simple adding and subtracting to find the answer.
Rac = R1 + R2, Rbd = R2 + R3, Rab= R1 Rcd = R3, which then can be written as
R1 + R2 = R2 + R3 + R1 – R3 which equals R1 + R2 = R1 +R2 which shows that they
are equal to each other. Since the voltage drop across each resistor is the same then
adding the voltage drops together as well doing ER1 + ER2 = ER1 + ER2 since ER3 cancels
itself out in the first equation given.
Part 2: Kirchoff’s Voltage Law
Theory:
Kirchoff’s Voltage Law states that the algebraic sum of all of the voltage drops around a
loop is 0. The key is that the “algebraic sum” means that voltage has a positive (+) or
negative (-) direction, depending on the actual direction of the voltage drop.
Figure 3:
Figure 4:
Data Table:
Circuit of Figure 3
Circuit of Figure 4
Voltage
Polarity Measured Voltage
Polarity Measured
+
12V
+
12V
E1
E1
1.44V
.91V
ER1
ER1
+
2V
3.25V
E2
E2
5.63V
3.54V
ER2
ER2
6.91V
4.32V
ER3
ER3
For the first part of part 2 (see figure 3) I connected a 2V DC power supply to the
schematic connecting the negative side of the power supply to the negative side of R1
making it in series with the rest of the resistors. I then measured the Input voltage of E1
which was 12V. Since this is an applied voltage, it has a positive polarity. Then I
measured the voltage drop of R1 and got 1.44V, since this is a drop in voltage the
polarity is a negative direction since the voltage is going down. Next, I measured the
voltage of E2, the added voltage, and got 2V. Since I am adding more voltage to the
circuit E2 is a positive polarity. Afterwards, I measured the voltage drop across R2 and
R3 and got 5.63V and 6.91V respectively. Since Kirchoff’s law states that the algebraic
sum adds up to 0 if we know the direction (polarity) of the voltage. For the first part I
found this to be true doing some arithmetic. + 12 – 1.44 + 2 – 5.63 – 6.91 = 0, this is
exactly what the law states.
For the second part of part 2 (see figure 4). I connected a 3.17V DC battery supply to the
schematic connecting the positive side of the power supply to the negative side of R1
making it in series with the rest of the resistors. I then measured the input voltage of E1,
which was 12V. Since this is an applied voltage, it has a positive polarity. Then I
measured the voltage drop across R1 and got 0.91V, and it is negative polarity because it
is a voltage drop. I proceeded to measure the voltage of the battery and found that it was
3.25V. Since the leads were reversed in the circuit, so was the polarity making it a
negative polarity instead of a positive polarity. Then I measured the voltage drop
(negative polarity) of R2 and R3 and got 3.54V and 4.32V respectively. Using the law
again we can take the sum of all these numbers and see what they total too. +12 - 0.91 –
3.25 – 3.54 – 4.32 = 0. Again this is exactly what the law states.
Part 3: Voltage Divider
Theory:
A voltage divider is a simple series circuit. Its purpose is to divide voltage so that a
smaller voltage is derived from a larger one. A voltage divider follows this formula:
Eout = Ein * (Rout / Rtotal)
Eout = Output voltage
Ein
= Input voltage
Rout = Resistor the output is measured across
Rtotal = total Resistance
Figure 5
Eout
R Calculated Eout measured
2 volts
124 ohms
2.02V
6 volts
620 ohms
6.04V
10 volts 3100 ohms
10.05V
For part 3 (see figure 3), I connected a decade box in series with a 620 ohm resistor and a
12 V DC power supply. By using the aforementioned formula I was able to calculate
what the resistance of the decade box (voltage divider) would need to be set to, to get the
desired voltage drop across it.
For the formula, I set it up as 2 = 12*(X/ (620+X)) where X is the resistance of the
decade box to get the desired output voltage I then multiplied both sides of the equation
by (620+X), which equaled 1240 + 2X = 12X (got this buy distributing the two across the
(620+X) and the 12 across the other X). Then I subtracted 2X from both sides and then
divided both sides by 10 to give me what X is equal to. In this case, I got 124 ohms. I set
up the same equation for each different desired outputs (6 = 12*(X/ (620+X), and 10 =
12*(X/ (620+X)) and solved for X each time to find the desired resistance and then
recorded my results.
Once I got my desired resistance figured out, I set the decade box to the first calculation
and measured the voltage drop which equaled 2.02V. Then I did this for the 620 ohm
setting and the 3100 ohm setting and recorded the voltage drop as 6.04V and 10.05V
respectively.
Reason why the voltage was not exact for the Eout and the measured Eout is because of the
tolerance that the resistor is set at which is + or – 5%, but the outcome is none the less
what I was looking for.
Answers:
1) The measured voltage across each resistor will add up to the applied voltage, the more
resistors in a series circuit the less the voltage will drop across an individual one. Also, it
is important to note that the smaller the resistor the smaller the voltage drops across it.
2) Ohm’s law states that the current remains the same in a series circuit. This can be
found to be true by going back to part one. For the circuit of figure one, the voltage
across resistor 1 (100 ohms) as 2.45V and the voltage across resistor 2 (390 ohms) as
9.56V. Ohms law states the E = I * R. Since E is know and R is known the equation can
be rewritten to find I. E/R =I. The current going through resistor 1 is calculated as
2.45/100 = 0.245 amps. Do the same thing for resistor 2 and it comes out to be .0245
amps as well. By going to figure 2 in part 1, one can prove ohms law as well even with
the extra 470 ohm resistor. Since the voltage drop across the resistors was smaller, the
applied current is also smaller but remains the same throughout the three resistors. Using
the same method as previously shown, the ampere going through each resistor is .0123
amps.
3) Kirchoff’s Voltage Law states that the algebraic sum of all of the voltage drops around
a loop is 0. . Using the polarity and the measured voltage of the circuit in figure 3,
Kirchoff’s Voltage Law can be proven: + 12 – 1.44 + 2 – 5.63 – 6.91 = 0, this is exactly
what the law states. Also, by using the measured voltage of the circuit in figure 4, the law
can be proven: +12 – 0.91 – 3.25 – 3.54 – 4.32 = 0. It is noted that voltage drop denotes a
negative polarity and a voltage increase denotes a positive increase.
4) Eac = Ebd + Eab -Ecd because in a series circuit the total resistance is the sum of all the
resistors, so in the case we can also use the Resistors to see how this works. It also allows
me to use simple adding and subtracting to find the answer.
Rac = R1 + R2, Rbd = R2 + R3, Rab= R1 Rcd = R3, which then can be written as
R1 + R2 = R2 + R3 + R1 – R3 which equals R1 + R2 = R1 +R2 which shows that they
are equal to each other. Since the voltage drop across each resistor is the same then
5) When a 100 ohm resistor is connected in parallel with the voltage divider (Decade
box) the total resistance of the circuit increases reducing the voltage going through the
divider. By observing what happened when the 100 ohm resistor was connected in
parallel with the divider, it increased the voltage drop across the 620 ohm resistor and
decreased the voltage drop across the divider.
Conclusion:
The lab clearly showed how voltage drops across a series circuit and the drop is equal to
the applied voltage. Also by using Kirchoff’s law it shows that the voltage going through
the circuit will be 0 because of the polarity of the circuit. Voltage drop negative polarity
voltage increase positive polarity. Take the algebraic sum of these and it will be 0. It
shows that all the voltage will be used up when it is done going through all the
components of the circuit. The decade box is a good tool to use to monitor what happens
to other components of the circuit if the resistance is changed. Through out the lab I
learned more about Kirchoff’s law and also to calculate the desired resistance of a decade
box if I want a desired voltage drop across it.