Experiment 5

Kirchhoff's Rules

 

Objective:

 

The objective of this experiment is to verify Kirchhoff's rules applied to a two-loop circuit.

 

Equipment:

A computer with Internet connection, a scientific calculator, paper, and pencil

 

Theory:

 

Ohm's law alone is not sufficient to solve for unknown currents in multi-loop circuits.  In such circuits, Kirchhoff's rules are used to solve for the unknown currents.  There are two rules:

 

Kirchhoff's Loop Rule (KLR)

Kirchhoff's Junction Rule (KJR)

 

a) KLR:

 

   This rule states that "the sum of voltage jumps and drops across the elements of any closed loop is zero."

   The voltages in a loop are either 1) due to an existing battery Vbat that we just write its voltage down, or  2) because of the current flowing in a resistor that we write the voltage across that resistor as the product RI.  The product RI is also a Voltage.  In writing a KLR, we just add the battery voltages and the RI voltages for all resistors in the loop and set the total equal to zero.  Of course, we need to be careful which ones are positive and which ones are negative as follows:

1) Rule for Batteries:  As we trace a closed loop in either direction, if we go from (-) to (+) across a battery, it is a voltage jump and we write +Vbat for that battery.  If we happen to go from (+) to (-) of it, it is a voltage drop and we write it as -Vbat . 

2) Rule for Resistors:  If our tracing direction opposes the assumed current I, we write +RI (a voltage jump) across resistor R, and if our tracing direction agrees with I, we write -RI (a voltage drop) across resistor R.

   The reason for the above is that when we swim against the flow of a river, we are trying to gain elevation, potential energy, or simply potential.  You know that water always flows from high elevation to low elevation.  Swimming against a water flow is an effort to gain elevation, or simply potential, and is positive.  Going from one end of a resistor to its other end against current I is like going to a higher potential or a higher voltage point of the loop that means a "voltage jump."  We then write the product RI with a (+) sign.  If our trace direction agrees with the assumed current I, it is like losing potential and the product RI deserves a (-) sign.

   Verify the correctness of Equations (1) and (2) for the 2-loop circuit shown below based on the assumed directions for currents I1 , I2, and I3 .

Figure 1

 KLR in Loop AGBCDA:    (1)

 -V2 +R3I3 +V1 +R2I1 +R1I1 = 0

 KLR in Loop AGBFEA:    (2)

 -V2 + R3I3 - R4I2 +V = 0.     

 KJR at Junction A:          

  - I1  +  I2  +  I3  =  0.   (3)     

Choosing Current Directions:  We don't know the true direction of the currents to begin with We simply assume a direction for the current in each branch.  If our assumption is correct, the answers we get at the end of our calculations will be positive.  If our assumption on a current direction is wrong, the calculations will yield a negative value for that current and we will know it at the end.

b) KJR:

This rule states that "The sum of currents going toward a junction is equal to the sum of currents leaving that junction."   In other words, the algebraic sum of currents to and away from a junction is zero. 

   A junction is a point of connection of 3 or more wires.  Equation (3) above is a KJR written for Junction A in Fig. 1.   I1 leaves Junction A and is given a (-) sign.   Both I2 and I3  go toward Junction A, they are given (+) signs.

A two-loop circuit has three branches.  For each branch, the current must be determined; therefore, there are three unknowns.  Three equations are needed to solve for three unknowns.  Two KLRs and one KJR will provide the three equations.

 

Procedure:

 

Click on http://www.physics.uoguelph.ca/applets/Intro_physics/kisalev/java/kirch1/index.html 

 

1) A 2-loop circuit containing 3 batteries and 3 resistors should appear.  If you keep clicking on the negative pole (blue) of any of the batteries, its voltage decreases until it becomes 1.0Volt.   Another click changes its polarity.  Clicking on its positive pole (red) increases its voltage to a maximum of 10.0 volts.  Experiment this for verification.   Also clicking on the top-end of any of the resistors, increases its resistance and clicking on the bottom end of it decreases its resistance.

 

2) As shown below, name the resistors as R1, R2, and R3 from left to right, respectively Each branch of the circuit in the applet has a battery in it.  Name the voltages of the batteries V1, V2, and V3 from left to right, respectively, as well.  On the applet, ignore the rectangular voltmeters that show the voltages across the resistors.  The oval ammeters are important for this experiment.  Name the current in the left branch, I1, in the  middle branch, I2, and in the right branch, I3 .  Of course the ammeters, A1, A2, and A3 measure currents I1, I2, and I3 for you.

 

3) Set the applet according to the data given in the first row of Table 1.

  

4) Double-check the values for battery voltages and resistances.

 

5) Read and record the currents I1, I2, and I3 as measured by the oval ammeters of the applet.  These are your measured values. 

 

DO NOT substitute these measured values in the equations you will write in the following steps. Also, ignore the +/- signs of these values.  Just write their absolute values in Table 1 as measured currents.

 

Procedure for Calculations:

 

Note:  I1, I2, and I3 must appear in the following equations as unknowns.

If you do not know how to use your calculator to solve simultaneous equations, the instructions for using TI 83, 84, 85, and 86 are given at the end of this manual.  Scroll down to see the instructions.

 

1) Assume one arbitrary direction for each of the currents  I1, I2 , and I3 in the left, middle, and right branches and show each by an arrow You need to draw the 2-loop circuit completely on paper and show your assumed direction for the current in each branch.  One possible assumption is shown in Fig. 2.  

   At the end of your calculations, you may get positive or negative numbers for the currents.  A positive value indicates that your assumed direction was correct.  A negative value shows that your assumed direction was wrong and that the real direction for that current is the opposite.

 

2) Apply Kirchhoff's Loop Rule (KLR) once to the left loop and once to the right loop to obtain 2 equations involving the 3 unknowns I1, I2, and I3.

 

3) Apply Kirchhoff's Junction Rule (KJR) to one junction (either Junction B, or Junction E).  This gives the 3rd equation you need to solve for the three accepted values of I1, I2, and I3.

 

4) Calculate a % error on each current.

 

5) Repeat the experiment for the remaining rows of Table 1.

Fig. 2

 

 

Data:   Given  and  Measured:

Table 1

 

Case

V1

Volts

V2

Volts

V3

Volts

R1

Ω

R2

Ω

R3

Ω

Currents: I1

Amps

I2

Amps

I3

Amps

                           
1 +

-

5.0 +

-

4.0 -

+

3.0 4.0 5.0 7.0 Measured      
 Accepted      
 % Error      
                           
2 +

-

3.0 -

+

6.0 -

+

8.0 5.0 4.0 9.0

 Measured

     
 Accepted      
 % Error      
                           
3 -

+

6.0 -

+

6.0 +

-

6.0 5.0 3.0 7.0  Measured      
 Accepted      
 % Error      

 

 

Calculations:

 

With the given values of V1, V2 , V3 , R1, R2 , and R3 , apply KLR and KJR to solve for I1, I2, and I3, and use these calculated values as accepted values.

 

Comparison of the Results:

 

Corresponding to every measured value, there is an accepted value.  Calculate a percent error on each current.

Conclusion: To be explained by students.

 

Discussion: To be explained by students.

 

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Solving System of Equations Using TI 83 and TI 84

 

To solve the following already rearranged (ordered) system that has 2 unknowns, we need a 2 x 3 matrix,

{ 3x + -2y = 12

6x +  4y  = -3

the A-matrix is:

3 -2

, and the B-matrix is 

12
6 4 -3

 

1) TI 83 and 84

To input and solve the matrix, 
1)  Press [2nd] [MATRX] (Note: On the TI-83, just press [MATRX]).
2)  Scroll to Edit.
3)  Press [1] to access matrix A.
4)  Input the dimensions [2] [ENTER] [3] [ENTER].   Note: See (*) below.
5)  Input the matrix entries, pressing enter after each value.
6)  Press [2nd] [QUIT] [2nd] [MATRX] (Press [2nd] [QUIT] [MATRX] on the TI-83).
7)  Scroll to MATH.
8)  Press [ALPHA] [B] [2nd] [MATRX] [1] [)] [ENTER].

     (With the TI-83, press [ALPHA] [B] [MATRX] [1] [)] [ENTER]).

The solution to the above system of 2 equations in 2 unknown should appear as 

x = 1.75 and y = -3.375.

(*) To solve for 2 unknowns, you will need a 2 x 3 matrix).

      To solve for 3 unknowns, you will need a 3 x 4 matrix).

      To solve for 4 unknowns, you will need a 4 x 5 matrix) and so on.

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Solving System of Equations Using TI 85 and 86

 

In the already ordered system:

3x + -2y = 12
6x +  4y  = -3 ,  the 1st row elements are: 3    -2  & 12

                    and the 2nd row elements are: 6     4  & -3.

 

1) Press [2nd] [TABLE]  to pop up the SIMULT for simultaneous equations.

2)  Enter the # of equs. to be solved.  For this example:  Number = 2  then Press [ENTER].

3) You will be asked for elements of the 1st row, here:  3   -2  &  12  Press [ENTER] after each entry.

4) You will be asked for elements of the 2nd row, here:  6   4  &  -3  Press [ENTER] after each entry.

5) Press [F5] to Solve.  You should get: x = 1.75 and y = -3.375.

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