## Resistance Calculations for Parallel CircuitsHopefully you've come from '...calculations for Series Circuits' ... and understand it. Parallel and series circuits have different characteristics - it is not possible to calculate resistance values using the same methods. Do Not Panic - this method is meant to be different! You can read about Current, Power, Voltage & Resistance here. To work out the resistor needed for your parallel circuit: Consider the circuit illustrated here. Let's assume the power source is 5 volts (the same as four nicely charged 'AA' rechargeables). There are three LEDs in the circuit, each needing 0.025 amps to light up at an optimum level. The Current that we need to flow from the battery is the sum of the currents: In a parallel circuit, the battery supplies the same voltage to each of the LEDs - it is as if each LED is connected across the battery separately. In our case we are using three blue LEDs that want a 3.4v supply. In our case we are using three blue LEDs that want a 3.4v supply. We can now calculate the value of the resistance needed to drop the supply from 5 volts...
Look at the resistances available (you can't get them in every value) and pick the exact value if it's available or the nearest higher value. In this case you'd want a 22Ω resistor. Finally we need to work out how much power the resistor is being asked to soak up, so we can use the correct type of resistor: Power (in watts) is calculated by multiplying current (i) by voltage (v). The power going into our circuit is Vsupply And our circuit is using 0.075 amps. We calculate that the power used will be 0.075 x 5 = 0.375 watts. Resistors are available in several wattages - we supplied 0.25 watt, 0.5 watt, 1, 2 and 6 watt types. As best practice you would be looking for a power rating of at least 10% more than the value calculated - 0.4 watts, then you'd get the next higher available type, ie 0.5 watt. |

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