Simple question about current draw

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In summary, a 10V, 100W motor needs 10A and has 1 Ohm of resistance while at 10 V, a 10 W motor needs 1 A and is 10 Ohms.
  • #1
kickflipper11
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Hello,

I've been working with some electrical stuff at my internship lately, but I don't have any real background in it, so I've been trying to learn what I can from the internet. One thing that I don't get involves Ohms law:

If I have a circuit powered by 10V with just a 1 Ohm resistor, the current flow will be 10A (I think). If that resistor is replaced by a motor, which would have a much larger resistance, the current flow will drop according to I=V/R. But common sense tells me that a larger motor would need more current for it to run. I'm sure I'm confusing the concepts. Basically, I can't mesh the idea that current draw will be higher to power larger motors and the idea that I=V/R.

I hope that makes sense, any response would really be appreciated, thanks!
 
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  • #2
I guess I'm not really clear what your issue is. Ohms law is it.

If you maintain a 10v supply (capable of supplying whatever current is needed) and you have a 1 ohm resister load then you get 1 amp of current and if you increase the resistance then you get less current and if you decrease the resistance you get more current.

A 10V DC motor generating 100Watts would need 10 amps (and so by implication would have a 1/10 ohm resistance.

It is possible to have a DC power supply that generates 1am with no problem but just can't hack 10amps, so with a given input voltage, stronger motors require a power supply that can drive more load.
 
  • #3
If I have a circuit powered by 10V with just a 1 Ohm resistor, the current flow will be 10A (I think)

In a theoretical diagram this is the case.

In a real world application a voltage source will have a current supply limit. If you find a lab power supply, most of them have the ability to be current limited, so if you set the limit to 1A and then tried to put 10V across a 1R resister, the supply would hit 1A and the voltage will drop back according to Ohms Law to 1V.
 
  • #4
Thanks a lot, that actually did answer my poorly phrased question. I had it in my head that a larger motor that had a higher rated power must have a higher resistance than a lower power motor, which was why I was confused because that wouldn't make sense with Ohm's Law. I didn't realize that a lower power motor would have a higher resistance. So a:

10 V, 100 W motor needs 10A and has 1 Ohm of resistance

while at 10 V, a 10 W motor needs 1 A and is 10 Ohms?
 
  • #5
kickflipper11 said:
while at 10 V, a 10 W motor needs 1 A and is 10 Ohms?

you got it
 
  • #6
Ok, I'm glad you cleared that up, thanks a lot it was really buggin me.
 

What is current draw?

Current draw refers to the amount of electrical current that is being drawn from a power source by an electronic device or circuit. It is typically measured in amperes (A) and can vary depending on the device's power requirements.

Why is it important to measure current draw?

Measuring current draw is important for several reasons. It helps determine the power requirements of a device, which is crucial for proper functioning and preventing damage. It also allows for troubleshooting and identifying any issues with the device's power consumption.

How can current draw be measured?

Current draw can be measured using a multimeter, which is a tool that measures electrical current, voltage, and resistance. It can also be measured using a current clamp, which is a device that clamps around a wire and uses a magnetic field to measure the current passing through it.

What factors can affect current draw?

Several factors can affect current draw, including the device's power requirements, the type and quality of power source, and any external components connected to the device. Temperature, voltage fluctuations, and circuit design can also impact current draw.

How can current draw be reduced?

Current draw can be reduced by using more efficient components, optimizing circuit design, and managing power usage. This can help extend the battery life of portable devices and reduce the cost of electricity for larger electronic systems.

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