Higher: Resistors in a network
When more than two resistors are connected in parallel the equation becomes:
\(\frac{1}{R}=\frac{1}{R}_{1} +\frac{1}{R}_{2} +\frac{1}{R}_{3} {...}\)
Example
The following resistor network is set up.
Calculate the total resistance of the network.
Answer
\(\frac{1}{R}=\frac{1}{R}_{1} +\frac{1}{R}_{2} +\frac{1}{R}_{3}\)
R1 = \({12}\Omega\)
R2 = \({18}\Omega\)
R3 = \({6}\Omega\)
\(\frac{1}{R}=\frac{1}{12} + \frac{1}{18} + \frac{1}{6}\)
\(\frac{1}{R}=\frac{11}{36}\)
R = \(\frac{36}{11}\)
R = \({3.27}\Omega\)
This means that the three individual resistors can be replaced by one resistor of \({3.27}\Omega\).
Adding resistors in parallel decreases the total resistance.
The current has a choice of paths and only has to pass along one branch of the circuit.
It does not pass through each resistor and the total resistance of a parallel circuit is always smaller than the smallest resistor.