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:
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.(Eq. 1)
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:
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?(Eq. 2)
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.
I_{1} + I_{2} + I_{3} 0 (Eq. 3)
Going around loop ABCF in a clockwise direction we get
going around loop DCFE in a counterclockwise direction we getV_{1} - I_{1}R_{1} I_{3}R_{3} = 0 (Eq. 4)
We now have three equations relating the three unknown currents so we can solve for each current.V_{2} - I_{2}R_{2} I_{3}R_{3} = 0 (Eq. 5)
Applying Kirchhoff's junction rule to the top branch point:
Solving for I_{3} we get equation (1):I_{1} + I_{2} = I_{3} (Eq. 6)
Applying Kirchhoff's loop rule to the left hand loop going around in a clockwise direction we getI_{3} = I_{1} + I_{2} (Eq. 7)
substituting for I_{3} from Eq. 7:10 - 3 I_{1} - 4 I_{3} 0 (Eq. 8)
and equation (2):10 - 3 I_{1} - 4 I_{1} 4 I_{2} = 0 (Eq. 9)
Now going around the right loop in a counterclockwise direction10 - 7 I_{1} - 4 I_{2} 0 (Eq. 10)
Again substituting for I_{3} we have2 - 2 I_{2}- 4 I_{3} = (Eq. 11)
And Equation (13) given below:2 - 2 I_{2} - 4 I_{1} - I_{2} = 0 (Eq. 12)
To solve for I_{1} we must somehow eliminate I_{2} from the equations. If we could arrange to make the coefficient of I_{2} the same in equations Eq. 10 and Eq. 13, we could subtract the two equations and eliminate I_{2}.2 - 4 I_{1} - 6 I_{2} 0 (Eq. 13)
Multiply Eq. 10 by 3 and multiply Eq. 13 by 2 and subtract the two equations:
30 - 21 I_{1} - 12 I_{2} 0 Eq. 10 X 3 (Eq. 14)4 - 8 I_{1} - 12 I_{2} 0 Eq. 13 X 2 (Eq. 15)
26 - 13 I_{1} = 0
13 I_{1} = 26
Substituting this value of I_{1} in Eq.13, we getI_{1} = 26/13 = A
The negative sign indicates that I_{2} is opposite to the direction assumed in the diagram. Substituting in Eq. 7, we get2 - 4×2 - 6 I_{2} =2 - 8 - 6 I_{2} = 0
- 6 - 6 I_{2} = 0
I_{2} = - 1 A
I_{3} = I_{1} I_{2} = 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
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: