[ 1 A, 4 Ω Ω , 1.1 A]

Sol.

R 0 = 2 Ω Ω , I 1 = 2.5A

I 2 =1.5A

V P = 19 – R 0 I

= 19 – (2) × (2.5)

V P = 14V

Similarly

V Q = 16V

⇒ I 3 = ⇒ I 3 = 1A

I 4 = = I 1 + I 3

⇒ I 4 = = I 1 + I 3

⇒ = 2.5+1

R = 4 Ω Ω

Applying KVL in FEQPF

– I 1 R 0 – I 3 R 0 + (I– I 1 ) ×R 0 = 0

I = I 3 + 2I 1 ... (i)

Applying KVL in AFEDCBA

V–I 1 R 1 – (I 1 – I 3 ) R X = 0

I 1 = .........(ii)

Applying KVL in AFCBA

V – (I–I 1 ) R 0 – (I–I 1 +I 3 ) R = 0

V – I (R 0 +R) +I 1 (R 0 +R) –I 3 R = 0

V– (I 3 +2I 1 ) (R 0 +R 1 ) +I 1 (R 0 +R) – I 3 R = 0 Using (i)

V – I 3 (R 0 +2R) –I 1 (R 0 +R) = 0

V– I 3 (R 0 +2R) – (R 0 +R) = 0 Using (ii)

The graph of the function is a hyperbola. Its special points are: at R x = 0, I 3 = –3.8 A ; at R x ; = 4I 3 = 0; at
R x = 32 Ω Ω , I 3 = 1A ; If R x → ∞ , I 3 tends to 19/16 = 1.1875A.