Physics / Unit 9: Electricity and Circuits
9.5 Series Circuits
Notes
In a series circuit, the parts are connected in one single loop.
- The current is the same through every part.
- Total resistance adds: $R_{\text{total}} = R_1 + R_2 + R_3 + \dots$
- The voltages across the parts add up to the battery voltage.
- If one bulb burns out, the loop is broken and all go out.
- Example: 2, 4 and $6 \text{ } \Omega$ in series with 24 V: $R = 12 \text{ } \Omega$, $I = 2 \text{ A}$, and the $6 \text{ } \Omega$ resistor has $V = 2 \times 6 = 12 \text{ V}$.
Practice
A $3 \text{ } \Omega$ and a $9 \text{ } \Omega$ resistor are connected in series to a 6 V battery. What is the current?
$R = 12 \text{ } \Omega$; $\displaystyle I = \frac{6}{12} = 0.5 \text{ A}$.
In the same circuit ($0.5 \text{ A}$ through $3 \text{ } \Omega$ and $9 \text{ } \Omega$), what is the voltage across the $9 \text{ } \Omega$ resistor?
$V = IR = 0.5 \times 9 = 4.5 \text{ V}$; the other $1.5 \text{ V}$ is across the $3 \text{ } \Omega$ resistor.
More identical bulbs are added in series to a battery. What happens?
Adding resistance in series lowers the current through every bulb.
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