A 230-V, single-phase domestic energy meter has a constant load of 4 A passing through it for 6 h at unity power factor. The meter disc makes 2208 revolutions during this period. What will be the energy consumed by the load if the meter disc completes 1240 revolutions?

Question

Question

This question was previously asked in

SSC JE EE Previous Paper 12 (Held on: 24 March 2021 Evening)

Answer 1

Concept:

In an energy meter, Meter constant (K) is given as,

\(K=\frac{R}{E}\)

And, \(R\propto E\)

Where,

R is the number of revolution and E is the energy consumed in kWh

And, E = VItcos ϕ

Calculation:

Given that,

Voltage = 230 V

Current = 4 A

cos ϕ = 1

Energy consumed in 6 hours is given as,

E₁ = 230 × 4 × 6 × 1 = 5520 Wh

R₁ = 2208

From above concept,

\(R\propto E\)

\(\therefore R_1E_2=R_2E_1\)

Given, R₂ = 1240

Hence,

\(E_2=\frac{R_2E_1}{R_1}=\frac{1240\times5520}{2208}=3100\ Wh\)

Hence, the Energy consumed in 1240 revolution is 3.1 kWh.