Right answer is (b) 5 A
Easiest explanation: 7I = V^2 + 2V …………………. (1)
Now, V = 10 – 1 ×I
Putting the value of V in eqt (1), we get,
&I = (10 – I) 2 + 2(10 – I)
Or, I = 100 + I2 – 20I + 20 – 2I
Or, I2 – 29I + 120 = 0…………………. (2)
∴ I = \(\frac{+29 ± \sqrt{29^2 – 4(120)}}{2} = \frac{29 ± 19}{2}\)
I = 5 A, 24 A
Now, I = 24 A is not possible because V will be negative from eqt (2)
∴ I = 5 A.