# Thread: Low current Circuits and Ohm's Law

1. ## Low current Circuits and Ohm's Law

I understand the basics of Ohm's Law, but I'm not sure of one thing. I realize that a resistor will lower the current in a circuit, and that a resistor (or something with resistance) turns some current into heat (or some other form of energy). So, will a resistor actually decrease the amp load on a battery? If not, how do you create a low current-drawing circuit?

2. v=ir. This is exactly what Ohms law is. Nothing more, nothing less. Voltage equals current times resistance. It's a simple math equation that will answer everything. But to answer simply, a resistor IS a load. A light bulb is a resistor that just happens to do something aside from generate heat. A computer is exactly the same thing. A big, fat, complex resistor. To the source of the current/voltage, a resistance is a resistance is a resistance.

Next step: if you take 2 resistors and run them in parallel

. . . /---R1---\
---< . . . . . . .>-----
. . . \---R2---/

The current is looking for the easiest way to get through the circuit. Much like a river seperating into 2 smaller rivers, the majority of the water will flow through the path of least resistance. In this case, yes, the higher resistance will have less current going through it. This is because it resists the current more than the resistor with a lower resistance.

Want to know how to get zero current? Look at v=ir -> i=v/r. Decrease the voltage to zero or add an infinite resistance. This basically means remove the battery (0 voltage) or disconnect the wires altogether (infinite resistance -- no path to take).

Ever take a battery and connect it with a wire? It gets really hot really quick. That's because the current draw is pretty much maximum. hold just a battery in your hand and the resistance is infinite. The battery will last you quite a while just sitting in your hand.

Now what was I saying? Oh yeah, to decrease the current draw on a battery, increase the resistance. But if you are messing with expensive and/or dangerous stuff, first you should do a few practice problems using Kirchoffs current law and Kirchoffs voltage law. You will understand everything a lot better with that. Simply sticking a resistor between your battery and your light bulb will only make the light bulb less bright.

What exactly are you trying to figure this out for anyways?

EDIT: Thought of another way to say it a lot easier. YES, adding a resistor (in series) to your circuit will decrease the current draw, but it also will decrease the voltage (which makes the light bulb dimmer as written above). In order to obtain the same voltages that you want, you will now have to increase the voltage, which will increase the current more. If the resistor you added still leaves you within the constraints you need, then perhaps you should look at a battery that has the lower voltage. Then you don't need to worry about the resistor at all.

3. It's real simple.
If you're working on a car and a crowbar falls across the + and - terminals of the battery, it'll spark, get hot and kill the battery fast because it has a very low resistance and draws a bunch of current. If there's nothing connecting the terminals, there's (theoretically) infinite resistance, which draws no current.
Sooo...
low resistance = high amperage
high resistance = low amperage

Hope that makes it a little more intuitive.

4. Thanks for the help, guys! The reason I was wondering is because I am modifying a Radioshack R/C car by using a circuit involving 2 relays, then connecting them to a larger, seperately powered motor. I want to decrease the load on the circuit that originally powered the motor, so I won't discharge the batteries as quickly.

5. Originally posted by Sniperboy
Thanks for the help, guys! The reason I was wondering is because I am modifying a Radioshack R/C car by using a circuit involving 2 relays, then connecting them to a larger, seperately powered motor. I want to decrease the load on the circuit that originally powered the motor, so I won't discharge the batteries as quickly.
The relays are liable to draw more on the battery than the motor.
Better to find a motor with a lower current draw.

Everything is relative. Lower the current rating and lower everything else too... such as motor rpm's and torque.

For the record the complete ohms law is...
E=IxR
R=E/I
I=E/R
P=IxE

So in your case, to lower the current draw on your R/C car, you have to reduce the power output by the motor.

6. I just thought of this: how about an op-amp to power my motor? Looking at the specs of some of them, they can't handle very much current, and I need to deliver 10 amps at max to my motor at around 9.6v. Is there a circuit possible that I can use for this high amperage operation?

7. Have you seen an op-amp? The ones I have seen are about the size of an eraser head, but half as thick. You will have to check it out, but I don't think they can handle much juice at all.

8. Ok, I'll stick with my original relay/diode circuit then.

9. When you are working with motors and such it is always better to change out the motor for a motor of different windings ratings etc because of losses associated with trying to lower the voltage to the unit. The motor won't make the same HP but it sure will last longer on a battery. I do rc cars also and I have high current and low current motors. some of the cheapest motors last the longest. So by having your relay circut turn on a low I unit and shut off the high I unit the batt will last longer. A relay with a low amp draw say up to 100 ma should be fine.

10. Originally posted by Sniperboy
Thanks for the help, guys! The reason I was wondering is because I am modifying a Radioshack R/C car by using a circuit involving 2 relays, then connecting them to a larger, seperately powered motor. I want to decrease the load on the circuit that originally powered the motor, so I won't discharge the batteries as quickly.
You realize that by reducing the overall current draw, you will effectivly reducing the motors capacity to propel the R/C car, ok? I.E. it will not be as fast as before, ok?
What you need to extend the batterylife is a switching unit. A unit that will reduce the output voltage to a desired level. That unit must only be connected to the motor since all other stuff on the car expects the 9,6V. All stuff on the car will be powered by 9,6V and the engine will recieve a defined maximum of, say, 7 volts. By cutting the voltage to 7, you have reduced the maximum power utput to about half. That might equal about half of maximum speed, I'm not sure.

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