Electricity · Lesson 6 of 9
Resistance of a System of Resistors
“Resistors cooperate in series and parallel, though they never agree on how to share the load.”
• Distinguish series and parallel resistor combinations from their connections. • Derive and use the equivalent resistance rule for series resistors. • Derive and use the equivalent resistance rule for parallel resistors. • Apply current and potential difference rules to circuit branches. • Reduce mixed resistor networks systematically. • Explain why electrical appliances are commonly connected in parallel.
Electrical devices rarely contain only one resistor. Several resistive components may share one path or provide separate paths between the same two points. Their arrangement controls the current, voltage distribution and total opposition offered to the source. A whole network can often be replaced by one equivalent resistance that draws the same total current at the same applied voltage.
The resistance of a single resistor that would draw the same total current as a given resistor combination across the same potential difference.
Resistors in Series
A connection in which resistors are joined end to end along a single unbranched current path.
Because there is only one path, the same current I passes through every series resistor. The source voltage is shared: V = V₁ + V₂ + V₃. Applying V = IR to the complete combination and to each resistor gives IRₛ = IR₁ + IR₂ + IR₃. Cancelling I produces the series rule.
The voltage drop across each series resistor is Vᵢ = IRᵢ. Since current is common, the larger resistance receives the larger share of the total potential difference. The individual drops must add to the source voltage, providing a useful calculation check.
Activity
Purpose: verify current and voltage rules in series. Connect three low-value resistors, a low-voltage battery, key and ammeter in one loop. Move the ammeter to several positions; its reading remains the same. Then connect a voltmeter across the complete group and separately across each resistor. Observation: V equals V₁ + V₂ + V₃. Conclusion: series current is common and potential differences add.
Problem
Find the equivalent resistance of 2 Ω, 3 Ω and 7 Ω in series.
- 1.All resistors lie in one path, so use the series rule.
- 2.Rₛ = 2 + 3 + 7 = 12 Ω.
- 3.The result exceeds 7 Ω, the largest individual resistance, as a series result should.
Problem
A 20 Ω lamp and 4 Ω resistor are in series across 6 V. Find current and each voltage drop.
- 1.Equivalent resistance: Rₛ = 20 + 4 = 24 Ω.
- 2.Current: I = V/Rₛ = 6/24 = 0.25 A.
- 3.Lamp voltage: V₁ = IR₁ = 0.25 × 20 = 5 V.
- 4.Resistor voltage: V₂ = 0.25 × 4 = 1 V.
- 5.Check: 5 V + 1 V = 6 V, equal to the source voltage.
Problem
Three resistors 5 Ω, 8 Ω and 12 Ω are in series across 10 V. Find the ammeter reading and voltage across 12 Ω.
- 1.Rₛ = 5 + 8 + 12 = 25 Ω.
- 2.The same current flows everywhere: I = 10/25 = 0.40 A.
- 3.Voltage across 12 Ω: V₁₂ = 0.40 × 12 = 4.8 V.
- 4.The other drops are 2.0 V and 3.2 V; their sum with 4.8 V is 10.0 V.
Resistors in Parallel
A connection in which resistors are joined between the same pair of junctions, providing separate current paths.
Every parallel branch begins and ends at the same two junctions, so each has the same potential difference V. Current divides among the branches and recombines: I = I₁ + I₂ + I₃. Using I = V/R for the entire group and each branch gives V/Rₚ = V/R₁ + V/R₂ + V/R₃. Cancelling V gives the reciprocal rule.
Activity
Purpose: verify current and voltage rules in parallel. Connect three resistors between the same two junctions of a low-voltage source. Measure voltage across the combination and each branch; the readings are equal. Place an ammeter in the main line and then in each branch. Observation: the total current equals the sum of branch currents. Conclusion: voltage is common and current divides.
Problem
Find the equivalent resistance of 6 Ω and 3 Ω in parallel.
- 1.Use 1/Rₚ = 1/6 + 1/3.
- 2.Write 1/3 as 2/6, giving 1/Rₚ = 3/6 = 1/2.
- 3.Therefore Rₚ = 2 Ω.
- 4.The result is below 3 Ω, the smallest branch resistance.
Problem
Resistors 5 Ω, 10 Ω and 30 Ω are in parallel across 12 V. Find branch currents and total current.
- 1.The same 12 V acts across every branch.
- 2.I₁ = 12/5 = 2.4 A; I₂ = 12/10 = 1.2 A; I₃ = 12/30 = 0.4 A.
- 3.Total I = 2.4 + 1.2 + 0.4 = 4.0 A.
- 4.Equivalent resistance Rₚ = V/I = 12/4 = 3 Ω, smaller than 5 Ω.
Problem
A parallel pair of 10 Ω and 40 Ω is in series with a parallel group of 30 Ω, 20 Ω and 60 Ω across 12 V. Find total current.
- 1.First group: 1/R′ = 1/10 + 1/40 = 5/40, so R′ = 8 Ω.
- 2.Second group: 1/R″ = 1/30 + 1/20 + 1/60 = 2/60 + 3/60 + 1/60 = 6/60, so R″ = 10 Ω.
- 3.The two equivalent groups are in series: R = 8 + 10 = 18 Ω.
- 4.Total current I = V/R = 12/18 = 0.667 A, approximately 0.67 A.
- 5.The order of reduction follows the actual junctions, not visual closeness on the page.
Why Parallel Connections Are Useful
Appliances require different currents but are designed for the same supply voltage. Parallel connection gives each appliance the full supply voltage and allows independent switching. If one branch opens, the others retain complete paths. A series arrangement would force one current through all appliances and a single failure would break the entire circuit.
When the aim is to obtain the highest possible equivalent resistance from a given set, connect all available resistors in series. To obtain the lowest possible equivalent resistance, connect them all in parallel. For a specified intermediate value, test small series or parallel subgroups and reduce each subgroup before combining it with the remainder.
Never add resistances merely because symbols appear side by side. Components are series only if their shared junction has no other branch. Components are parallel only if both of their ends connect to the same two junctions.
Quiz
Which quantity is the same through every resistor in series?
Which quantity is the same across every parallel branch?
What is the series equivalent of 4 Ω and 6 Ω?
What must be true of a parallel equivalent resistance?
Why are appliances connected in parallel?
Practice Problems
- Find the equivalent resistance of 4 Ω, 8 Ω and 12 Ω in series. Solution: Rₛ = 4 + 8 + 12 = 24 Ω.
- The series group in the previous problem is connected to 12 V. Find current and voltage across 8 Ω. Solution: I = 12/24 = 0.5 A. V₈ = IR = 0.5 × 8 = 4 V.
- Find the equivalent resistance of 4 Ω and 12 Ω in parallel. Solution: Rₚ = (4 × 12)/(4 + 12) = 48/16 = 3 Ω.
- Three resistors 2 Ω, 3 Ω and 6 Ω must give 4 Ω. Describe the arrangement. Solution: Put 3 Ω and 6 Ω in parallel: R = 18/9 = 2 Ω. Connect this 2 Ω equivalent in series with the 2 Ω resistor to obtain 4 Ω.
- Three resistors 2 Ω, 3 Ω and 6 Ω must give 1 Ω. Describe the arrangement. Solution: Connect all three in parallel. 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1, so R = 1 Ω.
Key Takeaways
• Series resistors share one current path and their resistances add. • Series voltage drops add to the source voltage. • Parallel branches share the same potential difference. • Total current equals the sum of parallel branch currents. • Parallel equivalent resistance is below the smallest branch resistance. • Mixed networks are reduced one clearly identified group at a time. • Parallel connection gives appliances full voltage and independent operation. • Current and voltage checks help verify numerical answers.