Example 1
Two resistors, 4.0Ω and 8.0Ω, connect in series across an ideal 12V source. Find current and drops.
Rtotal =12Ω; I=1212=1.0A. Drops are 1.0×4.0=4.0V and 1.0×8.0=8.0V, summing to 12V.
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Year 11 · SACE Stage 1
Series circuit analysis
Calculate current and voltage sharing in ideal series circuits.
Method and worked examples
A series circuit has one unbranched path, so steady current is the same through every component. Resistances add: Rtotal =R1+R2+ … . For an ideal source with negligible internal resistance and ideal connecting wires, I= Vsupply/Rtotal. The potential drops add to the source voltage because energy supplied per coulomb equals total energy transferred per coulomb around the loop.
Find total resistance first, then common current, then each drop with V= IR. Larger series resistance gets a larger share of the supply voltage at that current. Opening any point breaks the single path and stops steady current everywhere. These rules concern the circuit arrangement, not whether the physical drawing places resistors next to each other.
Example 1
Example 2
An ideal 9.0V source drives 0.30 A through two series resistors. One is 10Ω. Find the other.
Rtotal =IV=0.309.0=30Ω. The other resistor is 30−10=20Ω.
Watch out: There is no branch in which charge can accumulate steadily. Both carry the same current, though voltage drops can differ.