In the circuit shown in the figure given below, the value

In the circuit shown in the figure given below, the value of the current i will be given by

By KVL in loop (1)
5 – 1 × i ₁ – 1 × i₁ = 0

⇒ i₁ =	5	= 2.5 A
	2

∴ V_a = 1 × i₁ = 2.5 V ...........(1)
By KVL in loop (2)
4 V_ab = 3i + i

i =	4V_ab	= V_ab
	4

∴ i = V_ab ..........(2)
Hence in loop (2)
⇒ V_b = 1 × i
= 1 × V_ab
⇒ V_b = V_a – V_b

⇒ V_b =	V_a
	2

(Putting Va = 2.5 V from (1) we get

V_b =	2.5	= 1.25 V .......(3)
	2

From equations (1) and (3)
i = V_ab = (V_a – V_b)
= 2.5 – 1.25 = 1.25 A