Experiment 22

Experiment 22: DC Circuits - Kirchhoff’s Laws

Purpose:

To study the principles of DC networks and the use of Kirchhoff’s Rules.

Apparatus

1) a pre-wired circuit board with batteries, DC multimeter

2) a DC voltmeter electrical connectors

Theory: Definitions

1) A junction is a point where two or more wires connect.

2) A branch is a wire between two neighboring junctions. Each branch carries its own current.

3) A loop is a closed path, which starts and ends at the same junction.

Theory

The two Kirchhoff Laws which we will apply to our circuit are Kirchhoff’s Current Law and Kirchhoff’s Voltage Law.

A) Kirchhoff’s Current Law: The sum of all branch currents at a junction equals zero:

Σ Ii = 0 (1)

Currents have no intrinsic sign. The sign convention we use is that currents entering the junction are positive (+I), while all those leaving it are negative (-I). [(]

Example: Fig. 1 shows a junction where three branches (wires) meet. Each branch has its own current. At this junction, I1 enters and I2 and I3 leave: Σ Ii = +I1 - I2 - I3 = 0. This makes sense since I1 is supplying current to the junction while I2 and I3 are removing current from the junction.


B) Kirchhoff’s Voltage Law: The sum of all potential changes in a loop must be equal to zero. This tells us that the energy remains the same at the starting point no matter what path you take:

Σ Vi = 0 (2)

There are two types of potential changes in any circuit: 1) voltage differences designated by “V” ; and 2) electromotive forces (emf) given by batteries or power supplies designated by “e” .

When you apply the Voltage Law Equation (2) to a circuit, you trace a path around any closed loop. When you pass through a battery, write +e if your loop direction is from negative to positive because there is an energy gain. Write -V if you go through a resistor, capacitor, or inductor in the direction of the current since there is a loss of energy going through a resistor.

Example: Take a clockwise loop through the series circuit below (Fig. 2). The current direction is also clockwise. There are three current elements e, R1 and R2 that give voltages e, V1, and V2. Going in the clockwise direction, there is a voltage gain through the battery and voltage losses through the resistors. So, in this direction, Kirchhoff's voltage law gives us

+ e - V1 - V2 = 0. We can rearrange this to: e = V1 + V2 which shows that the two resistors (V1 and V2) use up all of the voltage supplied by the battery (e).

C) Ohm's law: This is the basic form of all voltage drops. The current, I, passes through the resistance, R, which causes a potential loss of V as electrons are pushed through it. The equation is one of the simplest in all of physics:

V = I · R (3)


Preliminary Procedure

a) Start by checking the zero reading of your ammeter with the terminals disconnected, and the zero reading of your voltmeter with the terminals shorted. If they do not read exactly zero, adjust them with the small screw in the center of the meter (under the needle indicator).

b) Your network board (shown with gaps closed by jumper wires) is shown in Fig. 3. Make sure that the batteries are mounted with polarities as shown in Fig. 3. Copy the circuit on your data sheet.

c) Record the values of resistances R1, R2, R3, and R4 on your data sheet. Measure and record the values of ε1 and ε2. (Note: ε1 is 3 volts and ε2 1.5 volts if the batteries are fresh.)

Procedure Part I: Kirchhoff's Current Law

d) When the gaps a-b, c-d, and e-f are closed with short wires called jumpers, there are three different currents in the network: I1in the left loop, I2 in the right loop, and I3 down the center.

e) To measure the direction of I1, use the 500mA scale on your ammeter, connecting it to the gap connectors a and b as shown in Fig. 4. below. This puts the ammeter in series with the resistors. Be careful to connect the + and – terminals of the ammeter properly. Make sure the other two gaps (c-d, and e-f) are closed with jumpers. If your reading is negative, switch the + and – wires. Once you have a positive reading, the current is going from the positive side to the negative side. Draw an arrow on your circuit diagram to record the direction of your current. If you’re not sure, call your instructor.


f) If everything is correct and the current is reasonable, switch the ammeter to the 50ma scale in order to get greater precision in your reading. Record the current I1 to 3 significant figures. Then disconnect the ammeter and reconnect gap a-b with a jumper.

g) To measure I2, remove the e-f jumper and connect the ammeter across the e-f gap, observing the same procedure as in (e) and (f) to find the magnitude and direction of the current. Then disconnect the ammeter and reconnect gap e-f with a jumper.

h) Repeat all this for I3 using gap c-d. Then disconnect the ammeter and reconnect gap c-d with a jumper.

i) Choose a junction (either c or g). Write the Kirchhoff’s Law Equation for this junction taking into account your measured directions of the currents. According to your measurements, is the equation (1) satisfied or not at this junction? If you are not certain, check with your instructor.

Procedure Part II: Kirchhoff's Voltage Law

j) On your data-sheet, prepare a Voltage Law table as shown. Attach a red connector to the plus (+) terminal of the voltmeter, and a black connector to the – terminal. Set your voltmeter to the 5-volt full scale. Remember voltage is measured by attaching the voltmeter to the appropriate connectors without disconnecting any wires in the circuit.

Kirchhoff's Voltage Law
VOLTAGE
BETWEEN
POINTS: / SIGN
(+ OR -) / MAGNITUDE
(VOLTS)
a-k
k-j
j-c
c-a
SUM

k) Connect the voltmeter to points in the circuit as indicated in this table as you go around the loop in the counter-clockwise direction. Remember the voltmeter will show you only the magnitude of the potential difference. The signs (plus and minus) must be determined by you by reading the sign conventions below carefully.

[Note: If you don’t get a reading, you must switch the positive and negative wires.] Measure and record all potential difference magnitudes and signs in the table.

VOLTAGE LAW SIGN CONVENTIONS
Resistor, Capacitor, Or Inductor
(Passive Circuit Component) / Battery Or Power Supply
( Active Circuit Component)
Loop in the same direction as current / - / Current loses energy as it passes through the component. / from - to + / + / Current gains energy as it passes through the component from negative to positive potential.
Loop in the opposite direction from current / + / from+ to - / -

l) According to your measurements, is the Kirchhoff Voltage Equation (2) satisfied or not? If you are not certain, check with your instructor.

AS SOON AS YOU FINISH THIS SECTION, DISCONNECT ALL GAP JUMPERS SO THAT THE BATTERY DOES NOT RUN DOWN.

Procedure Part III: The Modified Network

m) With all gaps open, connect a wire across the resistor R4. This effectively removes it from the circuit. Then reverse the polarity of emf, ε2 as shown in Fig. 5 .

n) When the gaps are closed, there will be 3 new values for I1, I2, and I3. (See Fig. 5.) Measure and record the currents in the Modified Network by the same procedure you used in Part I starting with the 500 ma scale as in (d); then proceed with (e) and (f).


BEFORE YOU LEAVE THE LAB:

a) Place ε2 into its original position.

b) Disconnect all jumpers from the circuit board.

c) Make sure you have recorded the values of your resistances R1, R2, R3, R4. (They are accurate to within 0.1%)

d) Make sure you have recorded the values of ε1 and ε2, from the voltmeter readings. (They are accurate to within 1.0%)

Lab Report

Part I: Verification of Kirchhoff’s Laws

1) Using your measured values of currents in Basic Network, display the actual value (in mA) of the left-hand-side of equation (1), when applied to junction g.

2) Using your measured magnitudes and signs of potential differences in Basic Network, display the actual value of the left-hand-side of equation (2) when applied to loop a-k-j-g-c-a (counter-clockwise about the left side loop).

Part II: Basic Network

3) Set up 3 equations with 3 unknowns (the currents) for the Basic Network: one for junction g, and the other two for loops a-k-j-g-c-a and i-g-c-f-i. You should include two loops equations and the current equation in terms of e’s, I’s and R’s.

4) Solve these equations by a method of your choice, but show clearly your method and include most of the steps in your calculation.

WARNING: Don’t use your measured values of currents in any of your calculations. You will use the recorded values of resistances and the emf's in order to determine the theoretical values of the currents.

NOTE: It is useful to solve your equations algebraically without inserting values for the resistances and emf's. Then it will be easier to calculate the currents for Part III: the Modified Network. For example, display the formula for I1, in the following form:


5) Display your results in Table 1 as shown. The discrepancies should be stated as the difference between the calculated and measured values, not in the % difference.

TABLE 1: BASIC CIRCUIT
CURRENT / CALCULATED
(mA) / MEASURED
(mA) / ABSOLUTE
DISCREPANCY
(mA)
I1
I2
I3

Question #1: Why are the internal resistances of the batteries not considered in this experiment?

Part III: Modified Network

6) Set up 3 equations with 3 unknowns for the modified network, as in (3) above.

Hint: You should now appreciate the importance of solving the equations algebraically, as in (3), because by reversing the sign of ε2 and setting R4 = 0 in (3) you automatically have the solution for the Modified Network.

7) Solve the modified equations. (Note: if you have done (3), just follow the hint above.)

8) Show your results for the modified network in a table similar to Table 1 above.

114

[(]* In calculations you can start with either sign. It makes no difference whether this is the correct direction or not since you will find the correct direction after you solve the equations. Solutions with the negative sign imply that the actual direction of the current is opposite to your assumed direction.