Resistors 2: Limiting Current 
Ohm’s Law
Resistance appears in the simplest form of a fundamental electronic relationship called Ohm’s law:
V = I ⋅ R 
where V is the potential difference or voltage drop measured in volts, I is the current in amperes (or amps), and R is the resistance in ohms. Ohm’s law says that voltage and current vary proportionally, where the constant of proportionality is the value of a resistor.
Here is a simple circuit schematic for applying Ohm’s law: The circle represents a constant voltage source with positive and negative terminals and the triangle at the bottom represents ground. The lines connecting these components represent wires or traces on a printed circuit board (PCB). This figure comes out of LTspice, a free program for simulating such circuits. We can build an actual circuit with a breadboard, a 9V battery, and 1K resistor.
Ohm’s law predicts that the current will be the ratio of voltage to resistance:
I = 

= 

= 0.009 amperes = 9mA (milliamperes) 
To see how to simulate this circuit in LTspice, see this introduction to LTspice. LTspice gives the same answer.
To see actual experimental results, I plan to write an introduction to using a multimeter.
Example: Inferring Current from Voltage
An immediate application of Ohm’s law is to measuring current somewhere in a circuit. To measure current through a circuit using a multimeter, you must break a connection and complete the circuit with the two probes of the multimeter. If you do not want to break a connection on a circuit board, you may still be able to measure the current indirectly using Ohm’s law. If you know a resistance and the voltage drop across that resistance, then you can calculate current as voltage divided by resistance.
With the voltage supply disconnected, use your multimeter to read the resistance of a resistor that is conducting the current you wish to measure. Alternatively, use the color bands on the resistor to determine its approximate resistance. With the voltage supply connected, use your multimeter to read the the voltage across the resistor. The ratio of the voltage reading in volts to the resistance in ohms is the approximate current in amperes.
This is one of the principles behind R.G.’s method for measuring the gain of transistors.
Example: Internal Resistance of a Power Supply
Taken at face value, Ohm’s law implies that a power supply that supplies a constant voltage will also supply an infinite current to a zeroresistance wire connecting the supply to ground: if V is constant, as R approaches zero I must approach infinity. But Ohm’s law and the constant voltage supply are just approximations. Actual power supplies have maximum current ratings that should, or must, be respected. If, for example, you hook up (or short) the positive and negative terminals of a real battery, the battery does not instantaneously deplete itself delivering an infinite current.
A better approximation to a real power supply is a constant voltage source in series with a resistor. The resistor represents the internal resistance of the power supply. A fresh 9V battery might supply 120mA so that its initial internal resistance would be
R = 

= 

= 75 ohms 
So replacing the 1K resistor in the schematic above with a 75 ohm resistor gives a better approximation to a 9V battery than a constant voltage source does.
Example: Resistor for an LED
We can also apply Ohm’s law to designing a guitar effects pedal with a light emitting diode (LED) to show when the guitar pedal effect is bypassed. The light will be on when the effect is in the signal path and off when bypassed. Here we focus on providing the LED with the appropriate power supply when the effect is swiched on.
I will describe diodes more generally on this site sometime in the near future. For this example, I will just say that our LED conducts current like a simple wire if there is at least 2V across its leads. The effect of the LED is that the voltage of the power supply appears to drop by 2V. So we can use Ohm’s law to predict the current in a simple circuit that powers an LED like (this one).
We need to predict the current through the LED because an LED burns out if it gets too much current. Common LEDs are designed to operate at 20mA and their current should not exceed 50mA. The initial current of a shorted 9V battery typically exceeds 100mA and many 9V DC power supplies are rated for several hundred mA. So if we simply hook up an LED to a 9V power source to an LED, the LED will probably just burn out.
However, by adding a resistor in series with the LED, we reproduce the simple Ohm’s law circuit above. The modified circuit is pictured in the schematic to the right, where an LED is followed by a resistor. We only have to adjust the voltage of the power supply. The presence of an LED does not affect the current but potential difference across the resistor drops by 2V from 9V. In this situation, Ohm’s law predicts how the level of the current varies with the level of the resistor:
I = 

The larger the resistance is, the smaller the current will be.
We can choose a value R for the resistor to limit the current to 20mA given V, which is 7V = 9V – 2V for a 9V power supply. For this reason, such a resistor is called a current limiting resistor. To get a 20mA = 0.020A current, Ohm’s law says
R = 

= 

= 350 ohms 
Because the standard ±5% resistor values in this neighborhood are 330 and 360, you would choose 360 to be on the safe side. Often, builders use much higher values, 4.7K for example, because battery drain is so much lower yet the LED is still bright enough.
Power
In my description of resistors, I reported that 1/4 watt resistors are recommended for building stompboxes. R.G. Keen gave a helpful explanation on Aron’s forum here. Power in watts equals voltage in ohms times current in amperes. Using Ohm’s law, formulas for power P are
P = V ⋅ I = I^{2} ⋅ R = 

The current limiting resistor for our LED is dissipating 7V at 20mA giving a power of 140mW. A 1/4W = 250mW can handle that, but not a smaller, 1/8W resistor. Admittedly, the current in other parts of the stompbox circuit are lower but one has to do such calculations to decide what power rating to use. And R.G. recommends that good practice is to use a rating roughly twice what you predict you will need.