Question

Kindly solve thisssssss

Answer 1

BEFORE STARTING WITH THIS PROBLEM....↓↓

Kirchhoff’s junction rule:

The sum of currents entering a junction is equal to the sum of currents leaving it.

Kirchhoff’s loop rule:

The sum of potential differences around a closed loop is zero

Using Kirchhoff’s loop rule on ABCDA loop,

80 - 20l₁ - 4I₂ = 0

8 - 2 I₁ - 4I₂ = 0

I₁ + 2I₂ - 4 = 0 → 1

Using Kirchhoff’s loop rule on CDEFC loop,

-40I₂  - 40 + 10(I₁-I₂) = 0

-40I₂  - 40 + 10I₁ - 10I₂ = 0

-50I₂ - 40 + 10I₁ = 0

I₁ - 5I₂ - 4 = 0 → 2

Subtracting (2) from (1), SO WE GET....

7I₂ = 0

∴ I₂ = 0

Substituting the value of I₂ in (1), we get....

I₁ = 4A

HENCE,

4A(amp) CURRENT IN 20Ω RESISTOR

NO CURRENT THROUGH 40Ω

HOPE IT HELPS YOU DEAR!!