Step 1: The efficiency (\( \eta \)) of an electric motor is given by: \[ \eta = \frac{\text{Output Power}}{\text{Input Power}} \times 100% \]
Step 2: Larger motors generally have higher full-load efficiency because: - Larger motors operate with lower relative losses. - Copper and iron losses are better managed. - Stray losses become negligible compared to output power.
Step 3: The efficiency of small motors (e.g., 1 kW or 5 kW) is lower due to: - Higher friction and windage losses. - Larger percentage of stray losses.
Step 4: Among the given options, the 100 kW motor has the highest expected full-load efficiency due to better energy conversion and reduced relative losses.
The bus impedance matrix of a 4-bus power system is given.
A branch having an impedance of \( j0.2 \Omega \) is connected between bus 2 and the reference. Then the values of \( Z_{22,new} \) and \( Z_{23,new} \) of the bus impedance matrix of the modified network are respectively _______.
When the input to Q is a 1 level, the frequency of oscillations of the timer circuit is _______.
The logic circuit given below converts a binary code \(Y_1, Y_2, Y_3\) into _______.
The bus admittance matrix of the network shown in the given figure, for which the marked parameters are per unit impedance, is _______.