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written 2.5 years ago by |

**Solution:**

**Kirchoff's current and voltage laws:**

In electrical circuits which consist of simple arrangement of resistors in series or in parallel with a given source of emf, all calculations can be carried out using Ohm's law.

But in more complicated circuits where there are junctions between more conductors and where more than one source of emf are active, then Kirchoff's laws can be useful for calculation purposes.

There are two laws called Kirchoff's current law and Kirchoff's voltage law.

**First law – Kirchoff's Current law or junction law:**

- “In any network of conductors, the algebraic sum of the currents meeting at any junction is zero”.

The convention applied in the junction rule is that the current flowing towards a junction is taken as positive and the current flowing away from the junction is taken as negative.

Let i1 , i2 , i3 , i4 and i5 be the current passing through the five conductors and meeting at the junction 'O'. Then according to Kirchoff's first law,

$ \begin{aligned} &i_{1}-i_{2}+i_{3}-i_{4}+i_{5}=0 \\\\ &\text { or } i_{1}+i_{3}+i_{5}=i_{2}+i_{4} \\ \end{aligned} $

That is, the sum of the currents flowing towards the junction is equal to the sum of the currents flowing away from the junction.

**Second law – Kirchoff's Voltage law or mesh law:**

- “In any closed electrical network, the algebraic sum of the products of current and resistance in each branch of a network is equal to the algebraic sum of the emfs in that network.”

The current direction may be assumed either clockwise or anti- clockwise. The convention applied is, if the current flows in the same direction in which we are considering the network or mesh, it is taken as positive and the current in opposite direction is taken as negative.

The emf which sends the current in the same direction in which the mesh is considered, it is taken to be positive and emf which sends the current in the opposite direction is taken to be negative.

**For the network BAGFB we have,**

$$ I_{1} R_{1}+I_{2}+I_{1} R_{3}+I_{3} R_{6}=E \\ $$

**For the network FBCDF we have,**

$$ I_{3} R_{6}-\left(I_{1}-I_{3}\right) R_{5}-\left(I_{1}-I_{3}\right) R_{4}=0 \\ $$