i want help in ohms law

Thread Starter

kavoshgar

Joined Aug 25, 2020
10
hi, I hope ur good and happy and happier... I a little confused in ohms law...for 2amp 20volt we need 40w resistor???!!!! (p=vi)

and one more question...if my i=3A and r=1000ohm then my voltage had to be 3000V?!!is this example another way to solve??

to this law power supply or battery have only voltage!and in reality, I have for example dc power 24v 3 Amp so that amp how effect in ohm law? how I can consider ps current to ohm law?
 

sghioto

Joined Dec 31, 2017
1,795
I have for example dc power 24v 3 Amp so that amp how effect in ohm law? how I can consider ps current to ohm law?
An example R=V/I so 24/3= 8 ohms. That would be the lowest value of resistance or load the supply is rated at.
SG
 
Last edited:

SamR

Joined Mar 19, 2019
2,888
What you seem to be missing is that the Power Supply design can provide up to 2A, your circuit will only draw the amperage it needs. You have a 24V supply and your load totals 2kΩ. So the amperage = V/R which is 24V/2kΩ=12mA. So out of 2 possible amps available, you are only using12mA of it. If the total amperage for the load is over 2A, you need a bigger power supply.
 

upand_at_them

Joined May 15, 2010
621
I a little confused in ohms law...for 2amp 20volt we need 40w resistor?(p=vi)
What you need depends on the circuit. Can you post the circuit you're working on? If you need to push 2 amps through a resistor and have it develop 20 volts across it, then yes the resistor would need to be able to dissipate at least 40 watts.

if my i=3A and r=1000ohm then my voltage had to be 3000V?
If you manage to push 3 amps through a 1000 ohm resistor then yes, the voltage across it is 3000 volts.
 
to this law power supply or battery have only voltage!and in reality
Not really. They have limits. A power supply has voltage and current limits. The limit might be a fuse. You should look up the HP DC power supply handbook, now keysight. It's free. There are various ways a power supply can limit current. A power supply can also be a constant current supply. An electronic load can be constant resistance and constant power.

A battery has an internal resistance, so it's modeled as an ideal voltage source and a resistor. An ideal voltage source has zero resistance.
 

Thread Starter

kavoshgar

Joined Aug 25, 2020
10
What you seem to be missing is that the Power Supply design can provide up to 2A, your circuit will only draw the amperage it needs. You have a 24V supply and your load totals 2kΩ. So the amperage = V/R which is 24V/2kΩ=12mA. So out of 2 possible amps available, you are only using12mA of it. If the total amperage for the load is over 2A, you need a bigger power supply.
thanks for answering...if I change the current of PS to 10amps and v=24 then what current pushed through resistor 2k?12ma?
 

Thread Starter

kavoshgar

Joined Aug 25, 2020
10
An example R=V/I so 24/3= 8 ohms. That would be the lowest value of resistance or load the supply is rated at.
SG
thanks a lot for fast answering...then how I cand know my resistor`s power?is this 3*24=72w?!! I don't think this resistor can be found by someone !so what is the solution ?
 

Papabravo

Joined Feb 24, 2006
14,831
The way to look at is:
  1. Choose any two of the three quantities: voltage, current, or resistance
  2. The remaining quantity is defined by the law
With respect to power supplies, although they have a Maximum Current Rating, that does not mean that they will supply that maximum current in all cases. They only supply that maximum amount of current to an appropriately chosen resistor.
 

Papabravo

Joined Feb 24, 2006
14,831
thanks a lot for fast answering...then how I cand know my resistor`s power?is this 3*24=72w?!! I don't think this resistor can be found by someone !so what is the solution ?
You don't chose a resistor for a specific power. If you need to dissipate 72 watts, then you search for a resistor that will dissipate more. For example you might find a resistor that will dissipate 100 watts. Using resistors to dissipate power is rarely done. Most resistors will dissipate 0.1 to 0.25 watts, rarely 0.5 watts.
 

Thread Starter

kavoshgar

Joined Aug 25, 2020
10
You don't chose a resistor for a specific power. If you need to dissipate 72 watts, then you search for a resistor that will dissipate more. For example you might find a resistor that will dissipate 100 watts. Using resistors to dissipate power is rarely done. Most resistors will dissipate 0.1 to 0.25 watts, rarely 0.5 watts.
if my resistor be 8 ohms then where I can found 72 or 100w resistor ?there is no other solution?cause if I found 100w resistor its be so expensive!
 

SamR

Joined Mar 19, 2019
2,888
When you look at the circuit (circuit analysis) you will have to calculate the V across each component. The voltage across the resistor, V/R, determines its A. V is the EMF available, A is the flow, and R is the flow control. The V*I(A) is the power, Watts that determines the Wattage size of the resistor. In a very simple way...
 

Papabravo

Joined Feb 24, 2006
14,831
if my resistor be 8 ohms then where I can found 72 or 100w resistor ?there is no other solution?cause if I found 100w resistor its be so expensive!
You can combine resistors in series or parallel to distribute the power over more parts. I'm curious about why you think this is a problem. If you need it then it doesn't matter how much it costs. If the solution costs too much, then you didn't really need it.
 

scorbin1

Joined Dec 24, 2019
55
The situations you use as examples seem to imply that you are sinking all of the available current into just this one resistor, and if that was true you would be absolutely correct. However unless your circuit is simply designed to waste power this is not very useful and probably not the actual case.

As SamR mentioned, you will have to calculate voltage across each component in the circuit, or at least the resistor in question. Just because you are supplying your circuit with 24V doesn't mean you will have 24V across your resistor, on the contrary it will almost assuredly be much lower than that. In series circuits the voltage is divided across all of the components in a series. A simple example would be a series of 2 resistors of the same value. If you apply 24V accross these resistors in series, you will effectively divide the voltage evenly between the two. Each resistor would see 12V accross it.

To calculate power requirements of a resistor in series with the rest of the circuit, you would need to measure, or calculate, the voltage drop accross the resistor, and the current through the circuit. Then those two numbers can be plugged into watts law and you can calculate the power dissipation necessary for that resistor. You don't have to find a resistor that matches that exact power rating, just find one that can handle AT LEAST that amount of power. Generally speaking you would prefer one that can handle a good bit more power than necessary to give yourself some head room.

If it is not in series then you have to take in to account all parallel components and their affect on the circuit and that's where this discussion could become very long. If this is the case you would need to research how to solve series/parallel circuits as even a generalized lesson in this is not very fitting for a forum reply.

If your calculation is calling for anything larger than a 0.5W resistor, you may need to rethink your circuit or recheck your work. It's very likely that you made a mistake in your calculation or there is a much better way of accomplishing your goal.
 
Last edited:

BobaMosfet

Joined Jul 1, 2009
1,288
hi, I hope ur good and happy and happier... I a little confused in ohms law...for 2amp 20volt we need 40w resistor???!!!! (p=vi)

and one more question...if my i=3A and r=1000ohm then my voltage had to be 3000V?!!is this example another way to solve??

to this law power supply or battery have only voltage!and in reality, I have for example dc power 24v 3 Amp so that amp how effect in ohm law? how I can consider ps current to ohm law?
P = IE is Watt's Law
E = IR is Ohm's Law

P = watts; E = volts; I = amps

This will help--

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3
 
ohmic relationships behave as V/I and is the constant R when V is in Volts and I is in Amps.

Unfortunately, your bringing in Power in a way that is not normally done. A positive power is power dissipated and a negative power is power generated. Turns out that your 5 MW power plant generates -5 MW of electricity.

Generally power is used as prerequisite for measureing consumption. i.e. the kWh or the killo-Watt-hour which is how we pay for electricity. Capacity of batteries is measured in say mAh or even Amp-hours.

When we select a resistor, we select one that can dissipate the power we anticipate using the P= equations.

When we want a heater, we do select a resistor that can dissipate the power we need as heat. In any event we then care about the heat we need.
 

atferrari

Joined Jan 6, 2004
4,166
The OP should get used to write with capitals and spaces where they belong. Reading his texts is a royal pain.

Units demand certain coherence. When a text is a mess and hard to follow you loose interest in replying.
 
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