Test: Measurement of Power & Energy- 2 - Electrical Engineering (EE) MCQ

Energy supplied = VI cosφ) x t x 10^-3 kWh
= 230 x 4 x 1 x 6 x 10^-3
= 5.52 kWh
Energy meter constant,

Energy consumed when the meter makes 1472 revolutions

Now, energy consumed
 VI cos φ x f x 10^-3 kWh = 230 x 5 x cost}) x 4 x 10^-3 kWh
= 3.68 kWh (As obtained above)
or, 
Hence, power factor of the load = 0.8

Here, cos φ = 1 (for d.c. watthour meter) Energy supplied to the energy meter in kWh

∴ Energy supplied to the energy meter per minute = 
Given, energy meter constant, K = 3275 revolutions/kWh
∴ Disc speed in rpm = Energy supplied per minute x Energy meter constant

An increase in temperature is accompanied by a rise in resistance of ail copper and aluminium parts.

Given, R_p = 400Ω,
R_S = 10,000 Ω
C_C = 20 pF
Self inductance of the pressure coil is given by
L_p = C_C x R_S²
= 20 x 10^-12 x (10⁴)²
= 20 x 10^-4 = 2 mH

Measurement of power factor with the help of two-wattmeter method can be done only if load is balanced. Hence, assertion is not true. Reason is a correct statement.

• Voltage: v(t) = 5 + 2 sin(ωt + 30°) + 10 sin(3ωt - 60°)

• Current: i(t) = 10 + 5 sin(ωt) + 8 sin(2ωt - 30°)

• Step 1: Calculate the instantaneous power, p(t)

  Instantaneous power is given by:

  p(t) = v(t) × i(t)

  Substitute the given expressions for v(t) and i(t):

  p(t) = (5 + 2 sin(ωt + 30°) + 10 sin(3ωt - 60°)) × (10 + 5 sin(ωt) + 8 sin(2ωt - 30°))

  Step 2: Find the average power

  The average power is the mean of the instantaneous power over a period T. For sinusoidal functions, the average power is calculated by considering only the constant terms (the DC components) and the terms with the same frequency (as the cross terms average to zero over a period).

• The DC components of voltage and current are:

  • Voltage DC component: 5

  • Current DC component: 10

• The power from these DC components is:

  P_DC = 5 × 10 = 50 watt

  For the sinusoidal terms, we consider the RMS (root mean square) values:

• For 2 sin(ωt + 30°) and 5 sin(ωt), the RMS values are:

  • Voltage RMS: 2

  • Current RMS: 5

• The power from the first sinusoidal component is:

  P_1 = 2 × 5 × cos(30°) = 5√3 ≈ 8.660 watt

• For 10 sin(3ωt - 60°) and 8 sin(2ωt - 30°), these have different frequencies, so they don’t contribute to the power directly.

• Thus, the total power consumed in the circuit is the sum of the DC power and the power from the first sinusoidal component:

  P_total = 50 + 8.660 ≈ 58.66 watt

  However, this value doesn't match the options provided, so the correct answer is:

  d) none of these

In an electrodynamometer type wattmeter:
• Both fixed and moving coils are air cored.
• Air friction damping is used due to low value of operating magnetic field.
• The moving coil is connected across the load.
• The fixed coil is connected is series with the load.

A wattmeter must be used for the measurement of power in a.c. circuit instead of merely an ammeter and a voltmeter because the power factor can’t be measured with the help of voltmeter and ammeter.
Wattmeter measures average active power not average reactive power. Hence, reason is not a correct statement.

In
In the above phasor diagram
I = load current and
I_P = Pressure coil current

We know that, measured power is

At leading power factor of load, angle (φ + β) is increased, therefore cos(φ + β) is reduced. Hence, wattmeter reads low at leading p.f, load.

We have:


or,

Phantom loading consists of supplying the pressure coil circuit (or moving coil circuit) from a circuit of required normal voltage, and the current circuit (or fixed coil circuit) from a separate low voltage supply. Hence, assertion is not true.
Reason is a correct statement.

Power loss in the voltage coil = 
instrument reading = 250 watt
∴ Actual load = 250 - 20
= 230 watt

Cripping is due to over compensation provided to the running friction. To avoid it holes are made on opposite side of the disc of the energy meter.

When power factor = 0.5 (φ = 60°), then one wattmeter will read zero. The load will be lagging in nature.
W₁ = V_LI_L cos (30 + φ) 
and W₂ = V_LI_L cos (30 - φ)
For φ = 60°, W₁ = 0 watt