Ohm's Law

What is Ohm's law?

Ohm's law (named for nineteenth century German physicist George Simon Ohm, who derived it)  is an equation describing the fundamental relationship between voltage, current, and resistance.  It is usually stated as: E = IR, or V=IR, where E or V = voltage, in volts. E stands for "electromotive force" which is the same thing as voltage, and I = current, in amps, and R = resistance, in ohms.

How is it used?

Ohm's law allows you to calculate the voltage drop across  a given resistor if you know the current flowing through the resistor.
Example:  How much voltage is dropped across a 10K ohm resistor which has 5mA flowing through it?  The  voltage dropped across the resistor is 50 volts, because E = IR = 5mA * 10K = 50 volts. (Note that "K" means 1000's of ohms, and "mA" means 1 x 10-3 amps).
The original E = IR equation can be manipulated to find any one of the three if the other two are known.   The three possible forms of the equation are:
(1)   E = IR
(2)   R = E/I
(3)   I = E/R
Example:  If you measure the current going into an 8 ohm speaker, and find it to be 2.5A RMS, the voltage across the speaker can be found by multiplying the current by the resistance, or E = IR = 2.5A*8ohms = 20V.

Example:  If you have a resistor of unknown value, and you measure 10V across it with your voltmeter, and then put an ammeter in series with the resistor to measure the current, and find it to be 100mA, the unknown resistance value could then be found by dividing the voltage across the resistor by the current through it, or R = E/I = 10V/100mA = 100 ohms.

Example:  If you have a 100K resistor, and you measure 100V across it with your voltmeter, the current flowing through the resistor can be found by dividing the voltage across the resistor by the resistance, or I = 100V/100K = 1mA.

The power equations

A related equation is used to calculate power in a circuit: P = EI, where P = power (in watts), E = voltage (in volts), and I = current (in amps).
Example: If you want to determine the output power of an amplifier, and measure 20V RMS across the load and 2.5A into the load, the power delivered by the amp to the load is: P = 20*2.5 = 50W.
As with the Ohm's Law equations, this equation can also be rearranged to solve for the other two quantities to produce three possible equations as follows:
(1)   P = EI
(2)   E = P/I
(3)   I = P/E.
Example:  If you have an amplifier that puts out 100W into a load, and you measure 2.5A into the load, the voltage across the load can be calculated as E = P/I = 100W/2.5A = 40V.
Example:  If you have an amplifier that puts out 50W into a load, and you measure 20V across the load, the amplifier is putting out a current of I = 50W/20V = 2.5A.
The power equation can also be combined with the first Ohm's law equation to derive a set of new equations.  Since E = IR, you can substitute IR for E in the power equation to obtain:
P = EI = (IR)*I,  or  P = I2R.
You can also find P if you know only E and R by substituting I=E/R into the power equation to obtain:
P = EI = E*(E/R), or P = E2/R.
These two equations can also be rearranged to solve for any one of the three variables if the other two are known.  This gives six possible equations relating power to voltage, current, and resistance as follows:
(1)   P = I2R
(2)   R = P/I2
(3)   I = sqrt(P/R)
(4)   P = E2/R
(5)   R = E2/P
(6)   E = sqrt(PR)

Example:  What power rating does a 10K resistor need to be if it is used as a decoupling filter in a power supply, and a total current of 10mA will be flowing through it? The answer can be found by using equation (1): P = I2R = (10mA)2 * 10K = 1 Watt.

Example:  What is the RMS voltage at the output of a 100W amplifier into a 16 ohm load?  The answer can be found using equation (6): E = sqrt(PR) = sqrt(100W*16ohms) = 40V RMS.

Example:  An amplifier is connected to an 8 ohm load.  The RMS voltage across it is measured using a DVM (digital volt meter) and found to be 15.5V right before clipping.  How much undistorted power can the amplifier put out into an 8 ohm load?  The answer can be found using equation (4): P = E2/R = 15.52/8 = 30 Watts.

Example:  How much current does a 100W amplifier put out into a 16 ohm load, an 8 ohm load, and a 4 ohm load?  The answer can be found using equation (3).  For the 16 ohm load, I = sqrt(P/R) = sqrt(100/16) = 2.5A   For the 8 ohm load,  I = sqrt(100/8) = 3.54A.  For the 4 ohm load,  I = sqrt(100/4) = 5A.
 

Summary

Ohm's Law is probably the single most important equation used in designing or analyzing a circuit.  These equations are useful for calculating many parameters in a guitar amplifier, including output power, voltage drops across resistors, current through resistors, power ratings of resistors, and many other things.

Following are all the equations that can be derived from Ohm's law and the power equations:

(1)     E = IR
(2)     R = E/I
(3)     I = E/R
(4)     P = EI
(5)     E = P/I
(6)     I = P/E.
(7)     P = I2R
(8)     R = P/I2
(9)     I = sqrt(P/R)
(10)   P = E2/R
(11)   R = E2/P
(12)   E = sqrt(PR)
It is not necessary to memorize all the equations, as they can be derived from the following two main equations, simply by either rearranging the equation to solve for one of the other two unknown variables, or by substituting the various derived equations from the first one into the second one:
(1)     E = IR
(2)     P = EI

Copyright © 1999,  Randall Aiken.  May not be reproduced in any form without written approval from Aiken Amplification.

Revised 2/18/14