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# Resistors in Series

## All resistors are connected end to end within a circuit. The resistance of the circuit is the total of their resistances.

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Resistors in Series

#### Resistors in Series

All resistors are connected end to end. Recall the river analogy: current is analogous to a river of water, but instead of water flowing, charge does. With resistors in series, there is only one river, so there is only one current. But since there is a voltage drop across each resistor, they may all have different voltages across them. The more resistors in series the more rocks in the river, so the less current that flows.

Key Equation
Rtotal=R1+R2+R3+\begin{align*}R_{total} = R_1 + R_2 + R_3 + \ldots\end{align*}

#### Example

A circuit is wired up with two resistors in series.

Both resistors are in the same ‘river’, so both have the same current flowing through them. Neither resistor has a direct connection to the power supply so neither has 20V across it. But the combined voltages across the individual resistors add up to 20V.

What is the total resistance of the circuit?

The total resistance is Rtotal=R1+R2=90Ω+10Ω=100Ω\begin{align*}R_{total}=R_1+R_2=90 \;\Omega+10 \;\Omega=100 \;\Omega\end{align*}

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=20V100Ω=0.20A\begin{align*}I_{total}=\frac{V_{total}}{R_{total}}=\frac{20\:V}{100\:\Omega}=0.20\:A\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=(0.20A)(20V)=4.0W\begin{align*}P_{total}=I_{total}V_{total}=(0.20\;A)(20V)=4.0\;W\end{align*}. So the Power Supply is outputting 4W (i.e. 4 Joules of energy per second).

How much power does each resistor dissipate?

Each resistor has different voltage across it, but the same current. So, using Ohm’s law, convert the power formula into a form that does not depend on voltage.

PP90ΩP10Ω=IV=I(IR)=I2R.=I290ΩR90Ω=(0.2A)2(90Ω)=3.6W=I210ΩR10Ω=(0.2A)2(10Ω)=0.4W\begin{align*}P&=IV=I(IR)=I^2R.\\ P_{90 \:\Omega} &= I^2_{90\:\Omega}R_{90\:\Omega}=(0.2\:A)^2(90\:\Omega)=3.6\:W\\ P_{10 \:\Omega} &= I^2_{10\:\Omega}R_{10\:\Omega}=(0.2\:A)^2(10\:\Omega)=0.4\:W\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 voltage is there across each resistor?

In order to calculate voltage across a resistor, use Ohm’s law.

V90ΩV10Ω=I90ΩR90Ω=(0.2A)(90Ω)=18V=I10ΩR10Ω=(0.2A)(10Ω)=2V\begin{align*}V_{90\:\Omega} &= I_{90\:\Omega}R_{90\:\Omega}=(0.2\:A)(90\:\Omega)=18\:V\\ V_{10\:\Omega} &= I_{10\:\Omega}R_{10\:\Omega}=(0.2\:A)(10\:\Omega)=2\:V\end{align*}

\begin{align*}^*\end{align*}Note: If you add up the voltages across the individual resistors you will obtain the total voltage of the circuit, as you should. Further note that with the voltages we can use the original form of the Power equation (P=IV)\begin{align*}(P=IV)\end{align*}, and we should get the same results as above.

P90ΩP10Ω=I90ΩV90Ω=(18V)(0.2A)=3.6W=I10ΩV10Ω=(2.0V)(0.2A)=0.4W\begin{align*}P_{90\:\Omega} &= I_{90\:\Omega}V_{90\:\Omega}=(18\:V)(0.2\:A)=3.6\:W\\ P_{10\:\Omega} &= I_{10\:\Omega}V_{10\:\Omega}=(2.0\:V)(0.2\:A)=0.4\:W\end{align*}

### Review

1. Regarding the circuit below.
1. If the ammeter reads 2A\begin{align*}2\;\mathrm{A}\end{align*}, what is the voltage?
2. How many watts is the power supply supplying?
3. How many watts are dissipated in each resistor?

2. Five resistors are wired in series. Their values are 10Ω\begin{align*} 10 \Omega \end{align*}, 56Ω\begin{align*} 56 \Omega \end{align*}, 82Ω\begin{align*} 82 \Omega \end{align*}, 120Ω\begin{align*} 120 \Omega \end{align*} and 180Ω\begin{align*} 180 \Omega \end{align*}.
1. If these resistors are connected to a 6 V battery, what is the current flowing out of the battery?
2. If these resistors are connected to a 120 V power supply, what is the current flowing out of the battery?
3. In order to increase current in your circuit, which two resistors would you remove?
3. Given the resistors above and a 12 V battery, how could you make a circuit that draws 0.0594 A?

1. a. 224V\begin{align*}224 \;\mathrm{V}\end{align*} b. 448W\begin{align*}448 \;\mathrm{W}\end{align*} c. 400W\begin{align*}400 \;\mathrm{W}\end{align*} by 100 Ω\begin{align*}100 \ \Omega\end{align*} and 48W\begin{align*}48 \;\mathrm{W}\end{align*} by 12 Ω\begin{align*}12 \ \Omega\end{align*}
2. a. 0.013 A b. 0.27 A c. 120Ω\begin{align*} 120 \Omega \end{align*} and 180Ω\begin{align*} 180 \Omega \end{align*}
3. Need about 202Ω\begin{align*} 202 \Omega \end{align*} of total resistance. So if you wire up the 120Ω\begin{align*} 120 \Omega \end{align*} and the 82Ω\begin{align*} 82 \Omega \end{align*} in series, you'll have it.

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