ABSTRACT
E1 D I1r1 C I1 C I2R i.e. 4 D 2I1 C 4I1 C I2, i.e. 6I1 C 4I2 D 4 1
From loop 2 of Fig. 13.5, and moving in an anticlockwise direction as indicated (once again, the choice of direction does not matter; it does not have to be in the same direction as that chosen for the first loop), gives:
E2 D I2r2 C I1 C I2R i.e. 2 D I2 C 4I1 C I2 i.e. 4I1 C 5I2 D 2 2
3 Solve Equations (1) and (2) for I1 and I2 2 ð 1 gives: 12I1 C 8I2 D 8 3 3 ð 2 gives: 12I1 C 15I2 D 6 4
3 4 gives: 7I2 D 2
hence I2 D 2/7 D 0.286A
(i.e. I2 is flowing in the opposite direction to that shown in Fig. 13.5)
From 1 6I1 C 40.286 D 4 6I1 D 4 C 1.144
Hence I1 D 5.1446 D 0.857A
Current flowing through resistance R is I1 C I2 D 0.857 C 0.286
D 0.571A
Note that a third loop is possible, as shown in Fig. 13.6, giving a third equation which can be used as a check: E1 E2 D I1r1 I2r2
4 2 D 2I1 I2 2 D 2I1 I2
D 2]
Problem 3. Determine, using Kirchhoff’s laws, each branch current for the network shown in Fig. 13.7
1 Currents, and their directions are shown labelled in Fig. 13.8 following Kirchhoff’s current law. It is usual, although not essential, to follow conventional current flow with current flowing from the positive terminal of the source
2 The network is divided into two loops as shown in Fig. 13.8. Applying Kirchhoff’s voltage law gives:
For loop 1: E1 C E2 D I1R1 C I2R2
For loop 2:
E2 D I2R2 I1 I2R3 Note that since loop 2 is in the opposite direction to current I1 I2, the volt drop across R3 (i.e. I1 I2R3) is by convention negative. Thus 12 D 2I2 5I1 I2 i.e. 12 D 5I1 C 7I2 2
3 Solving Equations (1) and (2) to find I1 and I2: 10 ð 1 gives: 160 D 5I1 C 20I2 3 2C3 gives: 172 D 27I2 hence I2 D 17227 D 6.37A
From (1): 16 D 0.5I1 C 26.37
I1 D 16 26.370.5 D 6.52A
Current flowing in R3 D I1 I2 D 6.52 6.37 D 0.15A
Problem 4. For the bridge network shown in Fig. 13.9 determine the currents in each of the resistors.
