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# Resistors in Parallel

## Many resistors branch from a single point. The resistance of the circuit is the sum of the inverse of each resistor.

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Resistors in Parallel

#### 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.

Key Equations

1Rtotal=1R1+1R2+1R3+\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 1Rtotal=1R1+1R2=190Ω+110Ω=190Ω+990Ω=1090Ω\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*} thus, Rtotal=90Ω10=9Ω\begin{align*}R_{total}=\frac{90\Omega}{10}=9\Omega\end{align*}

\begin{align*}^*\end{align*}Note: Total resistance for a circuit in parallel will always be smaller than smallest resistor in the circuit.

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

Use Ohm’s Law (V=IR)\begin{align*}(V=IR)\end{align*} but solve for current (I=V/R)\begin{align*}(I=V/R)\end{align*}.

Itotal=VtotalRtotal=20V9Ω=2.2A\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?

P=IV\begin{align*}P=IV\end{align*}, so the total power equals the total voltage multiplied by the total current. Thus, Ptotal=ItotalVtotal=(2.2A)(20V)=44.4W\begin{align*}P_{total}=I_{total}V_{total}=(2.2A)(20V)=44.4W\end{align*}. So the Power Supply outputs 44W (i.e. 44 Joules of energy per second).

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. P=IV=(VR)V=V2R\begin{align*}P=IV=\left(\frac{V}{R}\right) V=\frac{V^2}{R}\end{align*} Substituted I=V/R\begin{align*}I=V/R\end{align*} into the power formula. P90Ω=V290ΩR90Ω=(20V)290Ω=4.4W;P10Ω=V210ΩR10Ω=(20V)210Ω=40W\begin{align*}P_{90\Omega}=\frac{V^2_{90\Omega}}{R_{90\Omega}}=\frac{(20V)^2}{90\Omega}=4.4W; P_{10\Omega}=\frac{V^2_{10\Omega}}{R_{10}\Omega}=\frac{(20V)^2}{10\Omega}=40W\end{align*}

\begin{align*}^*\end{align*}Note: If you add up the power dissipated by each resistor, it equals the total power outputted, as it should–Energy is always conserved.

How much current is flowing through each resistor?

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

I90Ω=V90ΩR90Ω=20V90Ω=0.22AI10Ω=V10ΩR10Ω=20V10Ω=2.0A\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 10Ω\begin{align*}10\Omega\end{align*} resistor has the most current going through it. It has the least resistance to electricity so this makes sense.

\begin{align*}^*\end{align*}Note: If you add up the currents of the individual ‘rivers’ you get the total current of the of the circuit, as you should.

### Review

1. Three 82 Ω\begin{align*}82\ \Omega\end{align*} resistors and one 12 Ω\begin{align*}12\ \Omega\end{align*} resistor are wired in parallel with a 9V\begin{align*}9\;\mathrm{V}\end{align*}battery.
1. Draw the schematic diagram.
2. What is the total resistance of the circuit?
2. What does the ammeter read and which resistor is dissipating the most power?
3. Given three resistors, 200 Ω,300 Ω\begin{align*}200\ \Omega, 300\ \Omega\end{align*} and 600 Ω\begin{align*}600\ \Omega\end{align*} and a 120V\begin{align*}120\;\mathrm{V}\end{align*}power source connect them in a way to heat a container of water as rapidly as possible.
1. Show the circuit diagram
2. How many joules of heat are developed after 5 minutes?

1. b. 8.3W\begin{align*}8.3 \;\mathrm{W}\end{align*}
2. 0.8A\begin{align*}0.8\mathrm{A}\end{align*} and the 50 Ω\begin{align*}50 \ \Omega\end{align*} on the left
3. b. 43200J\begin{align*}43200\mathrm{J}\end{align*}.

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