P24 Manual
Return to List of Experiments for Physics 2122

Kirchhoff's Rules

(multiloop direct-current circuits)


To apply Kirchhoff's rules to a direct current network and compare calculated currents with measured currents.


The principles known as Kirchhoff's rules, in honor of the man who developed them, provide a means of obtaining enough independent equations to solve for the currents flowing in an electrical circuit.  There are many ways of stating Kirchhoff's rules but this experiment uses the formalism described below.

Kirchhoff's junction rule states that the algebraic sum of the currents at any branch point or junction in a circuit is zero.  Symbolically, we may write Equation 1, below, as the sum of n currents flowing into a junction:

                 (Eq. 1)
In applying this rule, currents flowing into a branch point are taken as positive, while currents flowing out of the branch point are taken as negative.  the opposite sign convention could also be used, but this would simply multiply the equation by (-1) and therefore would provide exactly the same information.

Kirchhoff's second rule states that the algebraic sum of the potential changes around any complete loop in the network is zero.  Again we may write Eq. 2, symbolically ,as the sum of the potential changes:

                (Eq. 2)
There V represents the potential change as we go across a particular circuit element.  If in going around the loop you go from the negative terminal of a source of e.m.f. (E) to the positive terminal, then the potential increases so that V = + E  neglecting the internal resistance of the source of e.m.f.).  In considering the potential change across a resistor, bear in mind that conventional current enters the high potential side of a resistor and leaves at the low potential side.  If you are going around a loop in the same direction as the current through a particular resistor then V = - I R.  of you are going around the loop in a direction opposite to the direction of the current through a particular resistor then V = + I R.  but how do you know the direction of the currents through the resistors before you solve the problem?

The answer is that in many cases you don't know, but it doesn't matter.  the first step in solving a problem using Kirchhoff's rules is to assume a direction for all the currents in the circuit.  Now apply Kirchhoff's rules and obtain enough equations to solve for all the unknown currents.  if your solution for a particular current comes out a positive number, it means that the current is in the direction you assumed.  If the solution for a particular current comes out as a negative number, it means that the actual current is opposite to the direction you assumed but that the magnitude you calculated is correct.  However, you must be sure to retain the minus sign in solving your equations for the other currents.

Applications of Kirchhoff's Rules

  1. Draw a diagram indicating all components and other quantities which are known.  As an example, consider the circuit shown below.
  2. indicate the direction of each current by an arrow.  As previously mentioned it does not matter what direction you assume for a particular current as long as you follow the sign conventions.
  3. Once you have assigned a direction to each current you may find it convenient to mark the high potential side of each resistor with a (+) sign and the low potential side with a (-) sign.  This will help in determining the sign of the potential changes as you apply Kirchhoff's loop rule.
  4. Apply Kirchhoff's junction rule to each branch point in the circuit.  actually if you have n  branch points you can only get (n - 1) independent equations.  In the case shown, we have two branch points so we get only one independent equation
  5. I1 + I2 + I3  0                     (Eq. 3)
  6. Now apply Kirchhoff's loop rule to as many closed loops of the circuit as are necessary to obtain enough independent equations to solve the problem.  In this case we have three unknown currents so we must obtain three independent equations.  We have already obtained one equation from Kirchhoff's junction rule so we must generate two more using Kirchhoff's loop rule.
  7. Going around loop ABCF in a clockwise direction we get

    V1 - I1R1 I3R3 = 0                (Eq. 4)
    going around loop DCFE in a counterclockwise direction we get
    V2 - I2R2 I3R3 = 0                (Eq. 5)
    We now have three equations relating the three unknown currents so we can solve for each current.
As an example of how to solve these equations consider the particular circuit shown below.

Applying Kirchhoff's junction rule to the top branch point:

I1 + I2 = I3                (Eq. 6)
Solving for I3  we get equation (1):
I3 = I1 + I2                       (Eq. 7)
Applying Kirchhoff's loop rule to the left hand loop going around in a clockwise direction we get
10 - 3 I1  - 4 I3    0          (Eq. 8)
substituting for I3  from Eq. 7:
10 - 3 I1  - 4 I1  4 I2  =  0             (Eq. 9)
and equation (2):
10 - 7 I1  - 4 I2    0                            (Eq. 10)
Now going around the right loop in a counterclockwise direction
2 - 2 I2- 4 I3  =                                  (Eq. 11)
Again substituting for I3  we have
2 - 2 I2 - 4 I1  - I2  =  0                    (Eq. 12)
And Equation (13) given below:
2 - 4 I1  - 6 I2    0                            (Eq. 13)
To solve for I1  we must somehow eliminate I2 from the equations.  If we could arrange to make the coefficient of I2 the same in equations Eq. 10 and Eq. 13, we could subtract the two equations and eliminate I2.

Multiply Eq. 10 by 3 and multiply Eq. 13 by 2 and subtract the two equations:

30 - 21 I1  - 12 I2   0     Eq. 10  X  3           (Eq. 14)

4 -  8 I1   - 12 I2   0       Eq. 13  X  2          (Eq. 15)

26 -  13 I1  =  0
13 I1  =  26
 I1  =  26/13  =  A
Substituting this value of  I1  in Eq.13, we get
2 -  4×2 -  6 I2

2 -  8 -  6 I2 =  0

-  6  - 6 I2 =  0

I2    =  - 1 A

The negative sign indicates that I2 is opposite to the direction assumed in the diagram.  Substituting in Eq. 7,  we get
I3   =  I1  I2  =  2 - 1  =  1 A


The apparatus used to study multi-loop direct-current circuits is shown below.

Arrange the apparatus as shown below. V1 represents a single dry cell and V2 represents two dry cells in series. R1 , R2 And R3 are decade resistance boxes.  The values you are to use for R1 , R2 and R3 should be attached to the desk where the experiment is located

With the circuit wired as shown, measure the terminal voltages V1 and V2 of the dry cells by connecting a voltmeter in a parallel across each voltage source.  Using the measured values of V1 and V2 and the nominal values of R1 , R2 , and R3 applying Kirchhoff's rules and obtain a sufficient number of equations to solve for I1 , I2 , and I3 .  Assume the directions of the currents are as shown on the diagram.  If any of your currents come out with a negative sign be sure you include this negative sign on your data sheet.  Complete the solutions for all equations before proceeding with the remainder of the experiment.

Now use an ammeter to measure each of the currents.  Break the circuit in each branch in turn and insert the ammeter in series with the resistor and reconnect the circuit.  Pay particular attention to the direction of current flow.  It if is opposite to the direction on the diagram, record it with a negative sign.

You should do the following:

  1. Use a pencil and paper to solve the equations obtained with Kirchhoff's Rules. Compare your measured currents with the calculated values.

  3. After you have done step 1 above, go to the program, circuit1, and let MATLAB solve the set of linear equations.

Points to Consider

  1. In using your voltmeter to measure V1 and V2 you assumed that when the voltmeter was connected in parallel with the dry cell the terminal voltage remained the same. This would be true for an ideal voltmeter which has effectively an infinite internal resistance.   However, a real voltmeter has a finite internal resistance and will thus draw some current from the cell and change its terminal voltage. How does this finite resistance of the voltmeter effect your measured voltages and therefore your calculated currents?
  2. In using your ammeter to measure currents you assumed that when the ammeter is inserted in series in the circuit the current remained the same. This would be true for an ideal ammeter which has zero internal resistance. real ammeter has some non-zero internal resistance. What effect does the resistance of the ammeter have on your measured currents.


  1. Run the program, circuit1, and report the results for the following cases:
    1. R1 = 0 ohm, R2 = 1 ohm, R3 1 ohm; E1 = 1 V, E2 = 1 V
    2. R1 = 1 ohm, R2 = 0 ohm, R3 1 ohm; E1 = 1 V, E2 = 1 V
    3. R1 = 1 ohm, R2 = 1 ohm, R3 1 ohm; E1 = 1 V, E2 = -1 V
    4. R1 = 0 ohm, R2 = 1 ohm, R3 1 ohm; E1 = 1 V, E2 = -1 V
    5. R1 = 1 ohm, R2 = 0 ohm, R3 1 ohm; E1 = 1 V, E2 = -1 V
  2. Use Equations (28-19), (28-20), and (28-21) in the textbook by Fishbane et al. and check the answers which you obtained in Exercise 1 above.
  3. Modify the program, circuit1, and solve the circuit shown in Figure (28-12) of the textbook by Fishbane et al. for R1 = R2 = R3 = 1 ohm, R4 = R5 = R6 = 2 ohm and E = 1 V.
  4. Modify the program, circuit1, to solve the circuit in Problem 22, Figure (28-31) of the textbook by Fishbane et al.

Open Laboratory

Links for Experiment P24

Courses for Scientists and Engineers