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Ohms law Voltage, Current and Resistance Ohms law connects the voltage across a device and the current flowing through it to its resistance. This presentation will teach you how to apply ohms law to some real situations. But before we start looking at formulas, let's remind ourselves about what voltage, current and resistance are. Let's consider a circuit as a tank of water with an out flow pipe. The amount of water in the tank, pushing down on the water below it, is the potential difference or voltage. The width of the pipe leaving the tank is the resistance (the smaller the pipe the bigger the resistance). The amount of water that flows out of the tank each second is the current. With this comparison, the rate at which the water flows out of the tank is given by the potential difference of the water in the tank and the width the outlet pipe. Ohms law states the equivalent observation for an electronic circuit. Ohms law tells us that the flow of current (I) in a circuit is given by the voltage (V) across the circuit divided by the resistance (R) of the circuit. Or in short This equation can be rearranged to give three formulas: Don't worry if your not good at maths, just remember the ohms law triangle. With this cover the letter you need and the position of the remaining two letters gives you the equation. So in the example below we have covered I to leave V/R Let's look at selecting a current limit resistor for an LED. Like diodes LEDs drop some voltage across them, this is typically 1.8 volts for a standard LED. If we are running off a 5 volt supply then the voltage across the resistor Vr = 5 - 1.8 = 3.2 V LEDs normally need about 10 mA (0.01 Amps) to operate, so we can work out the resistance using R =V/I So R = 3.2 / 0.01 = 320 ohms. Suppose we have a battery operated circuit that uses a push button switch as shown in the circuit diagram. When the switch is pressed current will be flowing through the resistor. If a lot of current flows through the resistor the batteries won't last long. If we are using a 9 volt battery and the pull up resistor is 1K we can calculate the current usage while the switch is pressed. I = V / R = 9 / 1,000 So I = 0.009 Amps = 9 mA