A 0.5HP induction motor was running at 240VAC. It was consuming 2.4 Amperes. Just when I connected a 4 microFarad capacitor in parallel with the motor's input, the current consumption reduced to 2.2 Amperes, but the motor's RPM was the same. How is this possible? Thanks.
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It has something to do with the power triangle.
The apparent power is rated VA (volt ampere). The reactive power is rated VAR (Volt ampere reactive) and real power is rated is watts
./| the hypotenuse is the apparent power, the bottom line is real power and the vertical
/_| line is the reactive power (i am sorry yahoo wont allow me to properly demonstrate the triangle.)
The reactive power from the inductor is positive from what I can remember and the reactive power created by the capacitor is negative. They subtract each other. If the reactive power decreases then the apparent power decreases. When one side of the triangle decreases, the hypotenuse decreases. The voltage supplied remains the same regardless but the current will decrease when the apparent power decreases. The apparent power is the direct multiplication of the voltage supplied and the current supplied. It is called apparent because it is power that the machine appears to draw. The machine will draw current during an interval of the cycle and then return it on the rest of the cycle. The net current in the line per cycle is initially 2.4 amperes. What the capacitor does is take some of the return current from the machine and store it. When the machine, needs the current it takes the current from the capacitor and the line. Since less current has to go back and forth on the line with the introduction of capacitor the current decreases 2.2 amperes.
0.5 hp is 373 watts. 373/240 = 1.55 amperes or 1.6 (I will use this number on my analogy)
If I was to explain this in non technical terms, think of the current as the number of cars in the road. If the auto repair shop takes in 20 car repairs in the morning but needs to have 16 cars in the afternoon so they have more space to work, it will have 20 cars come in the morning and 4 cars get sent back in the afternoon. The net traffic in the road to the auto shop is 20 cars coming in and 4 cars going back which is a total of 24 if summed up. Let me point out that the current component of the real power is 1.6 amperes so I would say that by the end of the day the auto shop finish 16 car repairs a day. Now lets have two parking spot or storage for two cars. The storage would be your capacitor in electrical terms. So instead of sending 4 cars back because of storage (space) problems, you are only sending 2 back since 2 can be stored in the shop. Net traffic 20 come in and 2 get sent back in the afternoon which is equal to 22. The introduction of storage reduces traffic in the road to the auto shop. Because of storage, the next day, the shop nows take in 18 cars plus 2 from storage which is a total of 20 (they like to have 20 incoming repairs each day) they send 4 cars back total of 22 traffic. Yes, I know, it's a very poorly managed shop.
The apparent power is rated VA (volt ampere). The reactive power is rated VAR (Volt ampere reactive) and real power is rated is watts
./| the hypotenuse is the apparent power, the bottom line is real power and the vertical
/_| line is the reactive power (i am sorry yahoo wont allow me to properly demonstrate the triangle.)
The reactive power from the inductor is positive from what I can remember and the reactive power created by the capacitor is negative. They subtract each other. If the reactive power decreases then the apparent power decreases. When one side of the triangle decreases, the hypotenuse decreases. The voltage supplied remains the same regardless but the current will decrease when the apparent power decreases. The apparent power is the direct multiplication of the voltage supplied and the current supplied. It is called apparent because it is power that the machine appears to draw. The machine will draw current during an interval of the cycle and then return it on the rest of the cycle. The net current in the line per cycle is initially 2.4 amperes. What the capacitor does is take some of the return current from the machine and store it. When the machine, needs the current it takes the current from the capacitor and the line. Since less current has to go back and forth on the line with the introduction of capacitor the current decreases 2.2 amperes.
0.5 hp is 373 watts. 373/240 = 1.55 amperes or 1.6 (I will use this number on my analogy)
If I was to explain this in non technical terms, think of the current as the number of cars in the road. If the auto repair shop takes in 20 car repairs in the morning but needs to have 16 cars in the afternoon so they have more space to work, it will have 20 cars come in the morning and 4 cars get sent back in the afternoon. The net traffic in the road to the auto shop is 20 cars coming in and 4 cars going back which is a total of 24 if summed up. Let me point out that the current component of the real power is 1.6 amperes so I would say that by the end of the day the auto shop finish 16 car repairs a day. Now lets have two parking spot or storage for two cars. The storage would be your capacitor in electrical terms. So instead of sending 4 cars back because of storage (space) problems, you are only sending 2 back since 2 can be stored in the shop. Net traffic 20 come in and 2 get sent back in the afternoon which is equal to 22. The introduction of storage reduces traffic in the road to the auto shop. Because of storage, the next day, the shop nows take in 18 cars plus 2 from storage which is a total of 20 (they like to have 20 incoming repairs each day) they send 4 cars back total of 22 traffic. Yes, I know, it's a very poorly managed shop.
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