Question

An ideal transformer with purely resistive load operates at 12 kV on the primary side. It supplies electrical energy to a number of nearby houses at 120 V. The average rate of energy consumption in the houses served by the transformer is 60 kW. The value of resistive load (Rs) required in the secondary circuit will be ________ mΩ.

Answer 1

$\frac{𝑉_𝑠}{𝑉_𝑝} = \frac{𝑁_𝑠}{𝑁_𝑝}$
⟹ $\frac{120}{12000} = \frac{𝑁_𝑠}{𝑁_𝑝}$
⟹ $\frac{𝑁_𝑠}{𝑁_𝑝} = \frac{1}{100}$ − − − (𝑖)
For an ideal transformer, input power = Output power
And power is given by 𝑃 = 𝑖𝑉
$𝑖_𝑝𝑉_𝑝 = 𝑖_𝑠𝑉_𝑠 = 60000𝑊$
$𝑖𝑝 = \frac{60000}{12000} = 5$
Now, 𝑅_𝑝 = \frac{𝑉_𝑝}{𝑖_𝑝}$$ = \frac{12000}{5} = 2400 Ω
$𝑅_𝑠 =\frac{𝑉_𝑠}{𝑖_𝑠} = \frac{120}{60000/120} = 120 × 120/60000 = 120/500 = 0.240Ω = 240 𝑚Ω$