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 \underline\hspace2cm m \Omega .

Alternating Current question

2023 · 6 Apr · Shift 1 · Q70

2023 · 6 Apr · Shift 1 · Q70

View written solutionFree

Correct answer: 240

Given data
- Primary voltage: $V_p = 12\,\text{kV}$
- Secondary voltage: $V_s = 120\,\text{V}$
- Power consumed by houses: $P = 60\,\text{kW}$
- Transformer is ideal, and load is purely resistive.
Use power relation on secondary side

Since the transformer is ideal, the output power equals the load power: $P = \frac{V_s^2}{R_s}$

Therefore, $R_s = \frac{V_s^2}{P}$
Substitute values

$R_s = \frac{(120)^2}{60\times 10^3}$

$R_s = \frac{14400}{60000} = 0.24\,\Omega$
Convert into milliohm

Since $1\,\Omega = 1000\,\text{m}\Omega$

we get $0.24\,\Omega = 240\,\text{m}\Omega$
Final answer

$\boxed{240}$

2023 · 6 Apr · Shift 1 · Q70

2023 · 6 Apr · Shift 1 · Q70

View written solutionFree

Correct answer: 240

Given data
- Primary voltage: $V_p = 12\,\text{kV}$
- Secondary voltage: $V_s = 120\,\text{V}$
- Power consumed by houses: $P = 60\,\text{kW}$
- Transformer is ideal, and load is purely resistive.
Use power relation on secondary side

Since the transformer is ideal, the output power equals the load power: $P = \frac{V_s^2}{R_s}$

Therefore, $R_s = \frac{V_s^2}{P}$
Substitute values

$R_s = \frac{(120)^2}{60\times 10^3}$

$R_s = \frac{14400}{60000} = 0.24\,\Omega$
Convert into milliohm

Since $1\,\Omega = 1000\,\text{m}\Omega$

we get $0.24\,\Omega = 240\,\text{m}\Omega$
Final answer

$\boxed{240}$