Lesson 3 of 4
Parallel circuits
I can state and apply the sum of parallel branch currents.Syllabus
Syllabus K223 / K224, 11(d). State and apply the sum of parallel branch currents.
I can state and apply the equal potential difference across parallel branches.Syllabus
Syllabus K223 / K224, 11(e). State and apply the equal potential difference across parallel branches.
I can recall and apply effective resistance for a number of resistors in series and in parallel.Syllabus
Syllabus K223 / K224, 11(f). Recall and apply effective resistance for a number of resistors in series and in parallel.
Make a guess. You are not marked.
One lamp is connected across a cell. A second identical lamp is now connected in parallel with the first. What happens to the current drawn from the cell?
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It doubles.
Each lamp still has the full cell p.d. across it, so each draws the same current as before. The cell now supplies both currents, twice as much in total.
In short
- In parallel, each branch has the full supply voltage.
- The branch currents add up to the main current.
- Adding a branch lowers the total resistance.
Parallel branches share the same two endpoints, so they have the same potential difference. Their currents add to the total current entering the group.
Follow the connection points, not the shape of the drawing. Two branches are parallel only when each begins and ends at the same pair of nodes.
The entering current divides at a junction and recombines at the other. Charge is conserved, so in steady conditions the total current entering equals the sum of the branch currents.
Each branch shares the same two endpoints
The source holds 6.0 V across the whole parallel network. Both finite-resistance branches carry current. Arrow lengths do not compare current magnitudes.
Reconnect the two resistors across the same two nodes
The equivalent resistance is 4/3 ohm, about 1.33 ohm. This is a different connection from the 6.0 ohm series circuit.
A changed circuit
The same two resistors, connected in parallel
With an ideal fixed 6.0 V source:
- Current in 2.0 Ω = 6.0/2.0 = 3.0 A.
- Current in 4.0 Ω = 6.0/4.0 = 1.5 A.
- Total current = 3.0 + 1.5 = 4.5 A.
Both branches carry current. The equivalent resistance is Vs/Itotal = 6.0/4.5 = 4/3 Ω, approximately 1.33 Ω.
Find the effective resistance
Each branch current is the common p.d. divided by its own resistance. Adding them gives Itotal = V/R1 + V/R2 + ... . Since Itotal = V/Rparallel, dividing by the common nonzero p.d. gives:
For exactly two resistors, this can be rearranged to Rparallel = R1R2/(R1 + R2). The general reciprocal relationship also works for three or more branches.
Three parallel branches
6 Ω, 12 Ω and 4 Ω
1/R = 1/6 + 1/12 + 1/4 = 2/12 + 1/12 + 3/12 = 1/2, so R = 2 Ω.
Across 6 V, the branch currents are 1.0 A, 0.50 A and 1.5 A. Their sum is 3.0 A, which agrees with 6 V / 2 Ω.
Try it step by step
Work it out in steps: Current in the smaller resistor in A, then Current in the larger resistor in A, then Total current in A, then Combined resistance in ohm.
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Problem 1.
Current in the smaller resistor in A: I1 = 6.0 / 3.0 = 2.0 A
Current in the larger resistor in A: I2 = 6.0 / 6.0 = 1.0 A
Total current in A: I = 2.0 + 1.0 = 3.0 A
Combined resistance in ohm: R = V / I = 6.0 / 3.0 = 2.0 ohm (check: 1/3 + 1/6 = 1/2)
Problem 2.
Current in the smaller resistor in A: I1 = 12 / 4.0 = 3.0 A
Current in the larger resistor in A: I2 = 12 / 12 = 1.0 A
Total current in A: I = 3.0 + 1.0 = 4.0 A
Combined resistance in ohm: R = 12 / 4.0 = 3.0 ohm (2 sf)
Problem 3.
Current in the smaller resistor in A: I1 = 10 / 5.0 = 2.0 A
Current in the larger resistor in A: I2 = 10 / 20 = 0.50 A
Total current in A: I = 2.0 + 0.50 = 2.5 A
Combined resistance in ohm: R = 10 / 2.5 = 4.0 ohm (2 sf)
For positive finite branch resistances, the parallel equivalent is smaller than the smallest branch resistance. The added path lets more total current flow at the same p.d.; it does not require all the current to choose only one path.
Open one branch
If these branches are connected directly across an ideal fixed-voltage source, opening one branch leaves the same p.d. across the remaining branches. Their currents stay the same, while total source current falls.
This prediction depends on the connection and supply condition. If another resistor is shared in series with the branches, their p.d. can change when a branch opens. The complete-circuit example shows why.
Check your understanding
Fixed 3.0 ohm and 6.0 ohm resistors are connected directly in parallel across an ideal 6.0 V source. What current does the source supply?
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3.0 A
Each branch has 6.0 V. Their currents are 6.0/3.0 = 2.0 A and 6.0/6.0 = 1.0 A, so the source supplies 2.0 + 1.0 = 3.0 A.
Practise an exam question
Write your answer first, then tick the mark points it makes.
Exam-style question
Two resistors are connected in parallel to a cell. State how the currents in the resistors are related to the current from the cell, and how the potential differences across them compare. [2]
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