
The power rating of an electric motor that draws a current of 3.75 A, when operated at 200 V, is nearly equals to:
A. $54 W$
B. $1 hp$
C. $500 W$
D. $800 W$
Answer
580.8k+ views
Hint: We know that power is defined as the product of current and voltage across the wire. With the help of the above definition we can easily determine the value of power. On just multiplying the given voltage with current we can easily determine the value of power. The SI unit of power is the watt or one joule per second.
Complete step by step solution:
According to initial data we have
Current, $I=3.75A$
Voltage, $V=200V$
We know that Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit. The SI unit of power is the watt or one joule per second.
Power is the product of current and voltage.
According to the above definition of power we have,
$P = V \times I$
Units of current and voltage is given in S.I. so need to convert,
So, on substituting the values of current and voltage we get
$\Rightarrow P = 200V \times 3.75A$
Since all are in S.I. unit then we have the power in S.I. unit i.e. is watt
On solving we get,
$P = 750W$
We know that Horsepower is a unit of measurement of power, or the rate at which work is done, usually in reference to the output of engines or motors. There are many different standards and types of horsepower.
One unit of horsepower, hp is equal to 750 W.
$P = 1hp$
$\therefore$ The power rating of an electric motor that draws a current of 3.75 A, when operated at 200 V, is equal to $1hp$. Hence the correct option is B.
Note:
The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft. The power involved in moving a ground vehicle is the product of the traction force on the wheels and the velocity of the vehicle. The power of a jet-propelled vehicle is the product of the engine thrust and the velocity of the vehicle. The rate at which a light bulb converts electrical energy into light and heat is measured in watts, the electrical energy used per unit of time.
Complete step by step solution:
According to initial data we have
Current, $I=3.75A$
Voltage, $V=200V$
We know that Electric power is the rate, per unit time, at which electrical energy is transferred by an electric circuit. The SI unit of power is the watt or one joule per second.
Power is the product of current and voltage.
According to the above definition of power we have,
$P = V \times I$
Units of current and voltage is given in S.I. so need to convert,
So, on substituting the values of current and voltage we get
$\Rightarrow P = 200V \times 3.75A$
Since all are in S.I. unit then we have the power in S.I. unit i.e. is watt
On solving we get,
$P = 750W$
We know that Horsepower is a unit of measurement of power, or the rate at which work is done, usually in reference to the output of engines or motors. There are many different standards and types of horsepower.
One unit of horsepower, hp is equal to 750 W.
$P = 1hp$
$\therefore$ The power rating of an electric motor that draws a current of 3.75 A, when operated at 200 V, is equal to $1hp$. Hence the correct option is B.
Note:
The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft. The power involved in moving a ground vehicle is the product of the traction force on the wheels and the velocity of the vehicle. The power of a jet-propelled vehicle is the product of the engine thrust and the velocity of the vehicle. The rate at which a light bulb converts electrical energy into light and heat is measured in watts, the electrical energy used per unit of time.
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