Higher

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 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}\OmegaR1 = {10}\Omega
R2 = {4}\OmegaR2 = {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}\OmegaR = {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.