#### Resistors in Parallel

All resistors are connected together at both ends. Recall the river analogy: current is analogous to a river of water, but instead of water flowing, charge does. With resistors in parallel, there are many rivers (i.e. the main river branches off into many other rivers), so all resistors receive different amounts of current. But since they are all connected to the same point at both ends they all receive the same voltage.

\begin{align*}\frac{1} {R_{total}} = \frac{1} {R_1} + \frac{1}{R_2} + \frac{1} {R_3} + \ldots\end{align*}

#### Example

A circuit is wired up with 2 resistors in parallel.

Both resistors are directly connected to the power supply, so both have the same 20V across them. But they are on different ‘rivers’ so they have different current flowing through them. Lets go through the same questions and answers as with the circuit in series.

What is the total resistance of the circuit?

The total resistance is \begin{align*}\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{90\Omega}+\frac{1}{10\Omega}=\frac{1}{90\Omega}+\frac{9}{90\Omega}=\frac{10}{90\Omega}\end{align*}

\begin{align*}^*\end{align*}

What is the total current coming out of the power supply?

Use Ohm’s Law \begin{align*}(V=IR)\end{align*}

\begin{align*}I_{total}=\frac{V_{total}}{R_{total}}=\frac{20V}{9\Omega}=2.2A\end{align*}

How much power does the power supply dissipate?

\begin{align*}P=IV\end{align*}

How much power is each resistor dissipating?

Each resistor has different current across it, but the same voltage. So, using Ohm’s law, convert the power formula into a form that does not depend on current. \begin{align*}P=IV=\left(\frac{V}{R}\right) V=\frac{V^2}{R}\end{align*}

\begin{align*}^*\end{align*}

How much current is flowing through each resistor?

Use Ohm’s law to calculate the current for each resistor.

\begin{align*}I_{90\Omega}=\frac{V_{90\Omega}}{R_{90\Omega}}=\frac{20V}{90\Omega}=0.22A \qquad I_{10\Omega}=\frac{V_{10\Omega}}{R_{10\Omega}}=\frac{20V}{10\Omega}=2.0A\end{align*}

Notice that the \begin{align*}10\Omega\end{align*}

\begin{align*}^*\end{align*}

### Interactive Simulations

### Review

- Three \begin{align*}82\ \Omega\end{align*}
82 Ω resistors and one \begin{align*}12\ \Omega\end{align*}12 Ω resistor are wired in parallel with a \begin{align*}9\;\mathrm{V}\end{align*}9V battery.- Draw the schematic diagram.
- What is the total resistance of the circuit?

- What does the ammeter read and which resistor is dissipating the most power?
- Given three resistors, \begin{align*}200\ \Omega, 300\ \Omega\end{align*}
200 Ω,300 Ω and \begin{align*}600\ \Omega\end{align*}600 Ω and a \begin{align*}120\;\mathrm{V}\end{align*}120V power source connect them in a way to heat a container of water as rapidly as possible.- Show the circuit diagram
- How many joules of heat are developed after 5 minutes?

### Review (Answers)

- b. \begin{align*}8.3 \;\mathrm{W}\end{align*}
8.3W - \begin{align*}0.8\mathrm{A}\end{align*}
0.8A and the \begin{align*}50 \ \Omega\end{align*}50 Ω on the left - b. \begin{align*}43200\mathrm{J}\end{align*}
43200J .