Final answer:
The efficiency of the 5-hp 120-V 41-A 1800r/min dc motor is approximately 75.81%. The calculation is based on the input power of 4920 watts and the output power of 3730 watts.
Step-by-step explanation:
To calculate the efficiency of the 5-hp 120-V 41-A 1800r/min shunt dc motor, we start by determining its input power and output power. The input power can be found by multiplying the supply voltage by the current (Pin = V * I). With the supply voltage at 120 volts and the current at 41 amperes, the input power is 4920 watts (120 V * 41 A).
The output power is the mechanical power produced by the motor, which for a 5-hp motor is 5 horsepower converted to watts (1 hp = 746 watts), equating to 3730 watts (5 hp * 746 W/hp).
To find the efficiency, we use the formula Efficiency (η) = (Pout / Pin) * 100. Inserting the input and output power values gives us an efficiency of (3730 W / 4920 W) * 100, which is approximately 75.81%.
It's important to note that we did not use the armature resistance and field resistance in this particular efficiency calculation as it is a simple power in vs. power out ratio, and losses in the motor are implied in the output power rating. However, those resistances would be used to calculate the power losses within the motor itself when considering a more detailed efficiency analysis.