Problem Statement: Find the current through the cell and tpd across 3rd wire in the given circuit. E1 =1.5v r1=2ohm E2=2v r2=1 ohm, have negative terminal joined by 6ohm and positive terminal by 4 ohm. Third wire of 8 ohm connected the mid point. Using Kirchoff's law explain in detail.

Solution:

The given circuit can be represented as follows:


Application of Kirchoff's Law:

1. Kirchoff's Current Law (KCL):
At any junction in an electrical circuit, the sum of currents entering the junction is equal to the sum of currents leaving the junction.
At point P, the current flowing towards the junction is I1 and the current flowing away from the junction is I2 + I3. Therefore, applying KCL at point P,
I1 = I2 + I3 ................. (1)

2. Kirchoff's Voltage Law (KVL):
In any closed loop of an electrical circuit, the algebraic sum of the potential differences is zero.
For the loop A-B-C-D-A, applying KVL,
- E1 + I1R1 + VBC - E2 - I2R2 = 0
- 1.5 + 2I1 + VBC - 2 - 2I2 = 0
- 2I1 + 2I2 + VBC = -0.5 .............. (2)

For the loop A-B-P-A, applying KVL,
- E1 + I1R1 - I3R3 - VAP = 0
- 1.5 + 2I1 - 4I3 - VAP = 0
- 2I1 + 4I3 = VAP - 1.5 .............. (3)

For the loop P-C-D-P, applying KVL,
- VAP + VBC - I2R2 - I3R3 = 0
- VAP + VBC - 2I2 - 8I3 = 0
VAP + VBC = 2I2 + 8I3 ............... (4)

Solving Equations:

From equation (1), we get
I1 = I2 + I3

Substituting this in equation (2), we get
- 2(I2 + I3) + 2I2 + VBC = -0.5
- I2 - I3 + VBC/2 = -0.25 ............... (5)

Substituting I1 = I2 + I3 in equation (3), we get
- 2(I2 + I3) + 4I3 = VAP - 1.5
2I3 = VAP - 0.5 - I2 ............... (6)

Substituting equations (5) and (6) in equation (4), we get
VAP + VBC = 2I2 + 8I3
VAP + VBC = 2I2 + 8[(VAP - 0.5 - I2)/2]
VAP + VBC = 4VAP - 2 - 6I