Find Amp of Circuit with Ohm's Law

[Pin Head]

Hi,
I'm trying to find the amp of a circuit. Okay I know a circuit has a resistivity, So for example if my circuit is made out of cooper which has a resistivity of 1.68 * 10 negative power of 8 and a voltage of 9 volts.So do I just use ohms law Volt / Resistant = Amp?

 

[Bobbywhy]

Yes, use Ohm's Law. Current equals Voltage divided by resistance. For a great explanation and built-in calculator, see:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmlaw.html

By the way, it's important to get your terminology and units correct

Current (I): units of Amperes
Voltage (E): units of Volts
Resistance (R): units of Ohms

 

[berkeman]

Hi,
I'm trying to find the amp of a circuit. Okay I know a circuit has a resistivity, So for example if my circuit is made out of cooper which has a resistivity of 1.68 * 10 negative power of 8 and a voltage of 9 volts.So do I just use ohms law Volt / Resistant = Amp?

Be sure to carry units along in your calculations -- it's important to have the units of the resistivity in order to get the correct answer for the resistance.

In this case, you would need to be sure to include the source resistance of your "9V" source. When you put a heavy load on a voltage source, its internal resistance needs to be part of your calculation (using Ohm's Law, yes).

 

[vk6kro]

Hi,
I'm trying to find the amp of a circuit. Okay I know a circuit has a resistivity, So for example if my circuit is made out of cooper which has a resistivity of 1.68 * 10 negative power of 8 and a voltage of 9 volts.So do I just use ohms law Volt / Resistant = Amp?

But first you will need to convert from resistivity to resistance.

For this, you need the formula:

Where
R = resistance of the metal sample.
ρ is the resistivity (in ohm -meters).
L is the length of the sample in the direction of current flow (in meters).
A is the cross sectional area of the sample (in square meters).

 

[alphysicist]

Hi there,

Yes, you are correct. Ohm's Law states that the current (in amps) flowing through a circuit is equal to the voltage (in volts) divided by the resistance (in ohms). So in your example, the current would be 9 volts divided by 1.68 * 10^-8 ohms, which equals 5.36 * 10^8 amps. Keep in mind that this is a very large amount of current and may not be realistic for a typical circuit. It's important to also consider other factors, such as the size and capacity of the circuit components, to ensure safe and efficient operation. Hope this helps!