In Series and Parallel Circuits you saw that a series circuit has a single path. Now we put numbers on it, using the resistances of its parts.
Why it works
With one path, every charge passes through every resistor, so the same current $I$ flows through each. Each resistor takes its share of the energy per coulomb, $V_i = IR_i$, and the shares must add up to what the battery gives: $V = IR_1 + IR_2 + IR_3 = I(R_1 + R_2 + R_3)$. The resistors act together like one resistor of $R_1 + R_2 + R_3$, which makes sense: it is like one long wire.
Key idea
Series: one current through all; the voltages add up to the battery's; $R_{\text{eq}} = R_1 + R_2 + \cdots$. Each resistor's share is $V_i = IR_i$, so the biggest resistor takes the biggest share of the voltage.
Worked examples
Worked example: Current and voltages
A $12\,\text{V}$ battery drives resistors of $2$, $4$ and $6\,\Omega$ in series. Find the current and each resistor's voltage.
- $R_{\text{eq}} = 2 + 4 + 6 = 12\,\Omega$, so $I = \dfrac{12\,\text{V}}{12\,\Omega} = 1.0\,\text{A}$.
- $V_1 = (1.0)(2) = 2\,\text{V}$, $V_2 = 4\,\text{V}$, $V_3 = 6\,\text{V}$.
- Check: $2 + 4 + 6 = 12\,\text{V}$, the battery's voltage.
Answer: $1.0\,\text{A}$; $2$, $4$ and $6\,\text{V}$
Worked example: A voltage divider
A $1.0\,\text{k}\Omega$ and a $2.0\,\text{k}\Omega$ resistor are in series across $9.0\,\text{V}$. What is the voltage across the $2.0\,\text{k}\Omega$ resistor?
- $I = \dfrac{9.0}{3000} = 3.0 \times 10^{-3}\,\text{A}$.
- $V_2 = IR_2 = (3.0 \times 10^{-3})(2000) = 6.0\,\text{V}$: two thirds of the voltage across two thirds of the resistance.
Answer: $6.0\,\text{V}$
Common mistake
Thinking the current gets smaller after each resistor, as if each one used some up. The current is the same everywhere in a series loop; what drops at each resistor is the potential. Check: an ammeter placed anywhere in a series circuit gives the same reading.
An ammeter measures current, so it goes in series, letting the current pass through it. A good ammeter has a very small resistance, so it barely changes the current it measures.
Because there is only one path, opening the circuit anywhere, or a single bulb burning out, stops the current everywhere.