Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains the same, so the copper loss will become (1/2)2 i.e 1/4 times of full load copper loss.
At full load copper Loss = I2R
At half load copper Loss = (I/2)2 × R = I2/4 × R
400 = I2/4 × R
I2R = 4 × 400
I2R = Full load copper Loss = 1600 W
Mutual inductance between the two coils is K = √L1.L2
K = √2 × 8
k = √16
k = 4 H
The inductance of the coil is given by the relation
L = Nφ/I
N = number of turns = 50
φ = flux = 200μWb
I = current = 8 A
L = 50 × 200 × 10−6 ⁄ 8
L = 1.25 mH
From the figure it can be concluded that the voltmeter reads 5 volts as shown in the figure below. Based on the voltmeter and ammeter readings in the measuring network, determine the value of the resistor R Here Current I = 1/2 A = 0.5 A Voltage V = 5 V R = V/I = 0.5/5 R = 10Ω
The luminous efficiency of an electric lamp is the ratio of the luminous flux (in lumen) emitted by the lamp to the electric power (watt) given to it i.e.
Luminous efficiency η = Luminous flux/Watt
Note:- Lumens is the unit of measurement for luminous flux, which is the total amount of visible light emitted by a source.