State Kirchhoff's rules of current distribution in an electrical network. Using these rules determine the value of the current `I_1` in the electric circuit of Fig.

Explanation

The circuit diagram can be redrawn as shown in Fig.

Here `I_3 = I_1 +I_2`
In mesh CABDA, we have
`-40.I_3 - 20.I_1 +40 = 0`
` - 40 (I_1 + I_2) - 20 I_1 + 40 = 0`
` rArr 60 I_1 + 40I_2 = 40` ....(ii)
and in mesh EABFE, we have
`-20.I_1 + 20I_2 + 80 = 0`
`rArr 20 I_1 - 20 I_2 = 80 ` ....(iii)
Multiplying (iii) by 2 and then adding with (ii), we get
`100 I_1 = - 120 `
` rArr I_1 = - 1.2 A `
The -ve sign shows that the direction of current is opposite to that shown in figure with an arrow mark.