A parallel circuit is one where the ends of all the devices are connected together. The resistors in the following circuit are in parallel. Just like current was constant in a series circuit, voltage is constant in a parallel circuit. This means that the voltage drop across all three resistors will be the same.
We still need to be able to calculate the current in this circuit and the current in each of the resistors. If the voltage drop is constant across all the resistors then we should be able to use Ohm’s law to calculate the current in each resistor. We’ll call each of these currents I1, I2 and I3 for the current drop in each resistor respectively.
I1 = V / R1
I2 = V / R2
I3 = V / R3
The total current in the circuit would have to be the sum of all these.
I = I1 + I2 + I3
This gives us…
I = V (1/R1 + 1/R2 + 1/R3)
This makes it a little bit harder to see how we can calculate the total resistance of this circuit so let’s try to substitute that back into our formula. By rearranging Ohm’s Law we get R = V / I.
So if we divide both sides of our previous equation by V we get…
I / V = 1/R1 + 1/R2 + 1/R3
I / V = 1 / R so…
1/R = 1/R1 + 1/R2 + 1/R3
This says that the inverse of the total resistance is equal to the sum of the inverses of all the parallel resistances. It’s a bit more complicated than the series circuit example but remember this is still just Ohm’s law.
Let’s try a real example.
V = 10 Volts
R1 = 100 Ω
R2 = 200 Ω
R3 = 250 Ω
We know that the voltage drop across each resistor is 10 Volts so let’s calculate the current in each one.
I1 = 10 V / 100 Ω = 0.1 A or 100 mA
I2 = 10 V / 200 Ω = 0.05 A or 50 mA
I3 = 10 V / 250 Ω = 0.04 A or 40 mA
So the total current in the circuit would be… 0.19 A or 190 mA
If we calculate the total resistance with the formula that we derived we’d get…
1/R = 1/100Ω + 1/200Ω + 1/250Ω
1/R = 0.019Ω
R = 52.63Ω
Let’s see if that get’s us the current that we originally calculated.
I = V / R
I = 10V / 52.63Ω
I = 0.19 A
So our equation works both ways.