Question

In the figure given below, the power loss in 1 Ω resistor using Norton’s Theorem is ________

(a) 9.76 W

(b) 9.26 W

(c) 10.76 W

(d) 11.70 W

Answer 1

Correct option is (b) 9.26 W

Easy explanation: Let us remove the 1 Ω resistor and short x-y.

At Node 1, assuming node potential to be V, (frac{V-10}{5}) + ISC = 5

But ISC = (frac{V}{2})

∴  (frac{V-10}{5} + frac{V}{2}) = 5

Or, 0.7 V = 7

That is, V= 10 V

∴ ISC = (frac{V}{2}) = 5 A

To find Rint, all constant sources are deactivated. Rint = (frac{(5+2)×2}{5+2+2} = frac{14}{9}) = 1.56 Ω

Rint = 1.56 Ω; ISC = IN = 5A

Here, I = IN (frac{R_{int}}{R_{int}+1} = 5 × frac{1.56}{1.56+1}) = 3.04 A

∴ Power loss = (3.04)^2 × 1 = 9.26 W.