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
Galvanized steel conductors do not corrode, and possess high resistance. Hence such Wires are used in telecommunications circuits, earth wires, guard wire, stray wire, etc.
Admittance (Y) is the reciprocal of the impedance of a circuit. Admittance of an AC circuit is analogous to the conductance of a DC circuit. The unit of Admittance is Simen or MHO Admittance = 1/Z simen Y = Conductance ± J Susceptance Or the Admittance can be written as Y = (G ± J B) Simen Now comparing the above equation by the given equation in the question i.e Y= a + jb ∴ a = G = Conductance
During the positive half cycle of the supply, diodes D1 and D2 conduct are forward biased and conduct current while diodes D3 and D4 are reverse biased and they act as an open circuit, the current flows through the load.
EXAMIANS