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
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.
When variable loss becomes equal to the constant loss, efficiency is maximum. Losses = Pi + Pc Since copper loss is a variable loss therefore Losses = Pi + Pi = 2pi Thus at a maximum efficiency of this transformer total loss = 150 x 2 = 300 W
EXAMIANS
