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The back emf constant and the armature resistance of a PMDC motor are 20mV/rad/s and 5Ω, respectively. When powered by a 10 V source, it draws 1.16 A current to drive a load. If power loss due to friction and other mechanical factors is 0.07 W, what is the efficiency of the motor at this operating point?

User HenningJ
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Final answer:

The efficiency of the PMDC motor at the given operating point can be calculated as the ratio of mechanical output power to electrical input power, and it is approximately 41.26% after considering the back EMF, armature resistance, and power loss due to friction.

Step-by-step explanation:

The efficiency of a motor can be calculated by the ratio of its mechanical output power to the electrical input power. In this case, the input power to the Permanent Magnet Direct Current (PMDC) motor can be calculated using the formula Input Power (Pin) = Voltage (V) × Current (I), which gives us Pin = 10 V × 1.16 A = 11.6 W. The output power can be found by taking into account the back electromotive force (back EMF), the armature resistance, and the power loss due to friction and other factors. The electrically developed power in the motor is given by Pdeveloped = I × (V - I × R), where I is the current, V is the supply voltage, and R is the armature resistance. Putting our values in, we get Pdeveloped = 1.16 A × (10 V - 1.16 A × 5Ω) = 11.6 W - 6.744 W = 4.856 W. To find the mechanical output power, we subtract the friction losses: Pmechanical = Pdeveloped - Friction Losses, thus Pmechanical = 4.856 W - 0.07 W = 4.786 W. The efficiency (Eff) is then calculated by Eff = Pmechanical / Pin, resulting in an efficiency of approximately Eff = 4.786 W / 11.6 W × 100% = 41.26%. Therefore, the efficiency of the motor at this operating point is about 41.26%.

User Joel G Mathew
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