Question

In the given circuit, find the current through each branch of the circuit and the potential drop across the 10 Ω resistor.

Answer 1

We have two loops e.g.,BCDE and ABEF
According to Kirchoff's rule , total potential across loop equals zero.
at junction entering currents equals to existing currrents.

apply Kirchoff's rule for loop EDCB ,
move from E---> D ----> C ---->B ---->E
4I₂ - 5 + 4 - 2I₁ = 0
⇒ 4I₂ - 2I₁ - 1 = 0 ------(1)

Apply Kirchoff's rule for loop EBAF
Move from E ---> B ---->A ---->F-----> E
-4I₂ + 5 - 10(I₁ + I₂) = 0
⇒-4I₂ + 5 - 10I₁ - 10I₂ = 0
⇒-14I₂ - 10I₁ + 5 = 0 ------(2)

Multiply 5 with equation (1) and subtract from equation (2),
-14I₂ - 10I₁ + 5 - 5(4I₂ - 2I₁ - 1) = 0
-14I₂ + 5 - 20I₂ + 5 = 0
-34I₂ + 10 = 0
I₂ = 10/34 = 5/17 A

And I₁ = (4I₂ - 1)/2 = 3/34 A

Now, potential across 10Ω = 10(I₁ + I₂) = 10(3/34 + 5/17)
= 10× 13/34
= 130/34 V