Calculating resistance - CCEAHigher example calculation

Learn how to calculate resistance in series and parallel circuits.

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Higher example calculation

Example

Calculate the total resistance of the network of resistors

Calculate the total resistance of the network of resistors.

This circuit contains a \({5}\Omega\) resistor in series with two resistors, \({6}\Omega\) and \({4}\Omega\), which are in parallel.

Start by calculating the combined resistance of the two parallel resistors.

\(\frac{1}{R}=\frac{1}{R}_{1}+\frac{1}{R}_{2}\)

R1 = \({6}\Omega\)

R2 = \({4}\Omega\)

\(\frac{1}{R}=\frac{1}{6} + \frac{1}{4}\)

\(\frac{1}{R}=\frac{5}{12}\)

R = \(\frac{12}{5}\)

R = \({2.4}\Omega\)

The network has been simplified to:

Simplified network of resistors

Now calculate the total resistance of the two resistors in series:

R = R1 + R 2

R = \({5}\Omega + {2.4}\Omega\)

R = \({7.4}\Omega\)

The total resistance of the network is \({7.4}\Omega\).

Example: Calculate total resistance

Calculate the total resistance of the network

Calculate the total resistance of the network.

Answer

The two \({4}\Omega\) resistors are in parallel with each other.

The two \({10}\Omega\) resistors are in parallel with each other.

The two parallel networks are in series with each other.

First, calculate the total resistance of each parallel network:

\(\frac{1}{R}=\frac{1}{R}_{1} +\frac{1}{R}_{2}\)\(\frac{1}{R}=\frac{1}{R}_{1} +\frac{1}{R}_{2}\)
R1 = \({4}\Omega\)R1 = \({10}\Omega\)
R2 = \({4}\Omega\)R2 = \({10}\Omega\)
\(\frac{1}{R}=\frac{1}{4} +\frac{1}{4}\)\(\frac{1}{R}=\frac{1}{10} +\frac{1}{10}\)
\(\frac{1}{R}=\frac{2}{4}\)\(\frac{1}{R}=\frac{2}{10}\)
R = \(\frac{4}{2}\)R = \(\frac{10}{2}\)
R = \({2}\Omega\)R = \({5}\Omega\)
\(\frac{1}{R}=\frac{1}{R}_{1} +\frac{1}{R}_{2}\)
\(\frac{1}{R}=\frac{1}{R}_{1} +\frac{1}{R}_{2}\)
R1 = \({4}\Omega\)
R1 = \({10}\Omega\)
R2 = \({4}\Omega\)
R2 = \({10}\Omega\)
\(\frac{1}{R}=\frac{1}{4} +\frac{1}{4}\)
\(\frac{1}{R}=\frac{1}{10} +\frac{1}{10}\)
\(\frac{1}{R}=\frac{2}{4}\)
\(\frac{1}{R}=\frac{2}{10}\)
R = \(\frac{4}{2}\)
R = \(\frac{10}{2}\)
R = \({2}\Omega\)
R = \({5}\Omega\)

The network has been simplified to:

The network has been simplified

Now calculate the total resistance of the two resistors in series.

R = R1 + R2

R = \({2}\Omega + {5}\Omega\)

R = \({7}\Omega\)

The total resistance of the network is \({7}\Omega\).