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Text lesson

Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits

Physics

Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits

Overview

Overview

🌳 Concept tree

  • Lesson 2 Resistor Combinations and Ohms Law for Closed Circuits
  • ├── Combining resistors
  • ├── Resistors in series
  • ├── Resistors in parallel
  • ├── Mixed networks
  • ├── Worked examples (new values)
  • ├── Ohm's law for a closed circuit
  • └── Terminal voltage
Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits · 1 / 9
Combining resistors

Combining resistors

  • Resistors are combined in only two basic ways, series and parallel.
  • The aim is always the same replace the group by one single equivalent resistance R that would draw the same current from the same source.
Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits · 2 / 9
Resistors in series

Resistors in series

Comparison aspectMain ideaKeywords
Purposeto obtain a larger resistance than any single resistorobtainlargerresistancesingle
How to measure the equivalent resistance experimentally
  • connect the resistors R₁
  • R₃ one after the other in a circuit that contains a battery
  • an ammeter
  • a rheostat (a variable resistor) and a switch. Close the circuit and adjust the rheostat until a convenient current I flows
  • read from the ammeter. Then use a voltmeter to measure the voltage across each resistor (V₁
  • V₃) and across the whole group (V). You will find that V = V₁ + V₂ + V₃
  • and that the ratio V / I gives the equivalent resistance
  • which equals the sum of the three resistances
connectresistorsaftercircuit
  • Facts for a series connection:
  • The same current I passes through every resistor (there is only one path, and charge cannot pile up).
  • The total voltage is the sum of the individual voltages V = V₁ + V₂ + V₃.
  • Derivation Since V = IR, V₁ = IR₁, V₂ = IR₂, V₃ = IR₃. Then IR = IR₁ + IR₂ + IR₃, and dividing by I:
  • R = R₁ + R₂ + R₃
Comparison aspectMain ideaKeywords
•
  • The equivalent resistance is larger than the largest resistor
  • the largest resistor dominates (it determines the total
  • as the book says)
equivalentresistancelargerlargest
For N equal resistors each of value R₀R = N R₀
The voltage divides in proportion to resistanceV₁ : V₂ : V₃ = R₁ : R₂ : R₃ (the bigger resistor takes the bigger share). This is the potential divider idea: V₁ = V × R₁ / (R₁ + R₂ + R₃)biggerresistortakesshare
Disadvantageif one component breaks (open circuit) the current stops in all of them. A string of old-fashioned decoration lamps all go out when one lamp burns out for exactly this reasoncomponentbreaksopencircuit
Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits · 3 / 9
Resistors in parallel

Resistors in parallel

Comparison aspectMain ideaKeywords
Purposeto obtain a smaller resistance out of several larger onesobtainsmallerresistanceseveral
How to measure the equivalent resistance experimentally
  • connect the three resistors side by side between the same two points
  • and place the group in a circuit with a battery
  • a rheostat and an ammeter in the main wire. Adjust the rheostat until a suitable main current I is read. Measure the voltage V across the group with a voltmeter and the current in each branch (I₁ in R₁
  • I₂ in R₂
  • I₃ in R₃) with the ammeter. You will find that I = I₁ + I₂ + I₃ and that V / I is smaller than the smallest resistance
connectthreeresistorsside
•Facts for a parallel connectionFactsparallelconnection
  • The same potential difference V is across every branch (all the branches start at one common point and end at another common point).
  • The main current splits I = I₁ + I₂ + I₃.
  • Derivation I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃, and I = V/R. Substituting into I = I₁ + I₂ + I₃ and dividing by V:
  • 1/R = 1/R₁ + 1/R₂ + 1/R₃
Comparison aspectMain ideaKeywords
For two resistors
  • R = R₁R₂ / (R₁ + R₂) (product over sum). This is quicker than adding fractions
  • use it whenever you have only two branches
R₁R₂productoverquicker
For N equal resistorsR = R₀ / N
•
  • The equivalent resistance is smaller than the smallest resistor. The smallest resistor carries the largest current (currents divide inversely with resistance: I₁ : I₂ = R₂ : R₁). The water analogy fits: the widest pipe (smallest resistance) carries most of the flow
  • and adding another pipe always makes it easier for the water to get through
  • so the total resistance falls
equivalentresistancesmallersmallest
Advantage
  • each branch works independently
  • which is why household appliances are connected in parallel: each gets the full mains voltage
  • and switching one off does not affect the others
branchworksindependentlyhousehold
•
  • For two resistors
  • the current in each branch can be found from the main current without finding the voltage: I₁ = I × R₂ / (R₁ + R₂) and I₂ = I × R₁ / (R₁ + R₂)
resistorscurrentbranchfound
Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits · 4 / 9
Mixed networks

Mixed networks

  • To find the equivalent resistance of a mixed network, replace each parallel or series group by its equivalent, working step by step from the farthest end towards the source.
  • To find individual currents and voltages, work backwards: total current → voltage across each group → current in each branch.
  • A reliable method:
  1. Identify the groups that are clearly series (the same current, with no junction between them) or clearly parallel (the same two end points).
  2. Replace the innermost group first and redraw a simpler circuit.
  3. Repeat until one resistor is left; that is R.
  4. Find the main current from I = V / R (or I = VB / (R + r) if the source has internal resistance).
  5. voltage across each equivalent resistor = I × its value; then split the current between parallel branches.
  6. Check by adding the branch currents and by checking that the voltages around the circuit add up to the battery voltage.
Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits · 5 / 9
Worked examples (new values)

Worked examples (new values)

  • Example 1 — series. Resistors of 10 Ω, 20 Ω and 30 Ω are in series with a 30 V battery of negligible internal resistance.
  • R = 10 + 20 + 30 = 60 Ω. I = 30/60 = 0.5 A in every resistor.
  • V₁ = 0.5 × 10 = 5 V, V₂ = 0.5 × 20 = 10 V, V₃ = 0.5 × 30 = 15 V; the sum is 30 V, equal to the battery emf.
  • Example 2 — parallel. The same three resistors are connected in parallel across the same battery.
  • Each has 30 V across it I₁ = 3 A, I₂ = 1.5 A, I₃ = 1 A, total 5.5 A.
  • 1/R = 1/10 + 1/20 + 1/30 = (6 + 3 + 2)/60 = 11/60, so R = 60/11 = 5.45 Ω. Check: 30 / 5.45 = 5.5 A.
  • Example 3 — mixed. A 4 Ω and a 12 Ω resistor are in parallel; this pair is in series with a 5 Ω resistor and a 16 V battery (r = 0).
  • Parallel pair 4×12 / 16 = 3 Ω. Total R = 3 + 5 = 8 Ω. I = 16/8 = 2 A.
  • Voltage across the pair = 2 × 3 = 6 V. So I(4 Ω) = 6/4 = 1.5 A and I(12 Ω) = 6/12 = 0.5 A (sum = 2 A).
  • Example 3a — a second series circuit. Three resistors of 8 Ω, 15 Ω and 27 Ω are joined in series across a 60 V supply (negligible internal resistance).
  • (a) R = 8 + 15 + 27 = 50 Ω.
  • (b) I = 60 / 50 = 1.2 A, the same in all three.
  • (c) V₁ = 1.2 × 8 = 9.6 V, V₂ = 1.2 × 15 = 18 V, V₃ = 1.2 × 27 = 32.4 V. Sum = 60 V ✔.
  • Notice that the 27 Ω resistor, being the largest, takes the largest share (54%) of the voltage.
  • Example 3b — a second parallel circuit. Resistors of 12 Ω, 20 Ω and 30 Ω are connected in parallel across a 24 V battery with negligible internal resistance.
  • (a) Each branch has 24 V I₁ = 24/12 = 2 A, I₂ = 24/20 = 1.2 A, I₃ = 24/30 = 0.8 A.
  • (b) The main current is I = 2 + 1.2 + 0.8 = 4 A.
  • (c) 1/R = 1/12 + 1/20 + 1/30 = (5 + 3 + 2)/60 = 10/60, so R = 6 Ω; check: 24/6 = 4 A ✔.
  • (d) R = 6 Ω is smaller than the smallest resistor (12 Ω), as it must be.
Comparison aspectMain ideaKeywords
Example 3c — a mixed circuit with the parallel group at the end.
  • A 20 Ω and a 5 Ω resistor are in parallel
  • this pair is in series with a 6 Ω resistor
  • and the combination is connected to a 15 V battery with negligible internal resistance
resistorpairseriescombination
Step 1parallel pair = 20 × 5 / 25 = 4 Ωparallelpair
Step 2R = 4 + 6 = 10 Ω
Step 3I = 15 / 10 = 1.5 A (this is also the current in the 6 Ω resistor)currentresistor
Step 4voltage across the pair = 1.5 × 4 = 6 Vvoltageacrosspair
Step 5
  • I(20 Ω) = 6 / 20 = 0.3 A
  • I(5 Ω) = 6 / 5 = 1.2 A
  • their sum is 1.5 A ✔
Example 3d — a series group inside a parallel group.
  • A 6 Ω and a 12 Ω resistor in series form one branch
  • this branch is in parallel with a 9 Ω resistor
  • the pair of branches is in series with a 4 Ω resistor
  • across a 20 V battery of negligible internal resistance
resistorformbranchpair
Step 1the series branch = 6 + 12 = 18 Ωseriesbranch
Step 2parallel with 9 Ω = 18 × 9 / 27 = 6 Ωparallel
Step 3
  • R = 6 + 4 = 10 Ω
  • so I = 20 / 10 = 2 A
Step 4
  • voltage across the parallel part = 2 × 6 = 12 V
  • the 4 Ω resistor takes 8 V (12 + 8 = 20 ✔)
voltageacrossparallelpart
Step 5
  • I in the 18 Ω branch = 12 / 18 = 0.667 A
  • I in the 9 Ω resistor = 12 / 9 = 1.333 A
  • sum = 2 A ✔
branchresistor
Step 6the voltage across the 6 Ω resistor is 0.667 × 6 = 4 V and across the 12 Ω resistor is 8 Vvoltageacrossresistor
Example 3e — equal resistors.→Five identical 10 Ω resistors are connected (i) in series and (ii) in parallel. (i) R = 5 × 10 = 50 Ω. (ii) R = 10 / 5 = 2 Ω. The ratio of the two combinations is 25 : 1.
Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits · 6 / 9
Ohm’s law for a closed circuit

Ohm’s law for a closed circuit

  • A real source has an internal resistance r (the resistance of the electrolyte, plates, wires, etc.
  • inside the cell).
  • The current passes through the external resistance R and the internal resistance r in series.
  • The emf equals the total work per coulomb round the whole circuit (outside plus inside the source), so
  • VB = I R + I r = I (R + r)
  • I = VB / (R + r)
This is Ohm’s law for a closed circuitthe current equals the total emf divided by the total (external + internal) resistance of the circuit.
In wordstotal emf over total resistance.
Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits · 7 / 9
Terminal voltage

Terminal voltage

  • The voltage across the terminals of the source while it delivers current is

V = VB − I r (and also V = I R)

Comparison aspectMain ideaKeywords
The term I r is the "lost volts"
  • it is the energy per coulomb used up inside the source
  • so it is not available to the external circuit
energycoulombusedinside
•If the circuit is open (I = 0) the voltmeter across the source reads V = VB. This gives another definition: the emf of a source is the potential difference across its terminals when no current is flowing in the circuit.circuitopenvoltmeteracross
•As the external resistance R is increased (for example by moving a rheostat), I falls, I r falls, and the terminal voltage rises towards VB. As R is decreased (towards a short circuit) the current becomes large (VB / r), and the terminal voltage drops towards zero.externalresistanceincreasedexample
•A voltmeter of very high resistance across the terminals reads (almost exactly) the emf only when the circuit is openvoltmeterveryhighresistance
•Internal resistance can be found from r = (VB − V) / I.Internalresistancefound
Experiment idea (the book’s Fig 1-5)connect a cell, a rheostat, an ammeter and a switch in series, with a voltmeter across the cell. With the switch open the voltmeter reads the emf. Close the switch and the reading drops to V = VB − I r. Increasing the rheostat resistance lowers I and brings the reading back towards VB, which shows how V and I are related by a straight line.connectcellrheostatammeter
Short circuitwhen the external resistance is zero, I = VB / r. Because r is small the current is very large; this can overheat and damage the source and the wires, and it is why fuses and circuit breakers are used.externalresistancezeroBecause
Chargingif a source is forced to carry current in the opposite direction (current entering its positive terminal, as when a battery is being recharged by a stronger source), the terminal voltage is V = VB + I r, greater than the emf.sourceforcedcarrycurrent
Example 4.A cell of emf 6 V and internal resistance 0.4 Ω is connected to an external resistor of 11.6 Ωcellinternalresistanceconnected
•I = 6 / (11.6 + 0.4) = 0.5 A. Terminal voltage V = 6 − 0.5 × 0.4 = 5.8 V (= 0.5 × 11.6)Terminalvoltage
Example 4a — a second case.A battery of emf 9 V and internal resistance 0.5 Ω supplies a 4 Ω resistorbatteryinternalresistancesupplies
Step 1total resistance = 4 + 0.5 = 4.5 Ωtotalresistance
Step 2I = 9 / 4.5 = 2 A
Step 3V = VB − I r = 9 − 2 × 0.5 = 8 V (check: I R = 2 × 4 = 8 V ✔)check
Step 4lost volts = 1 Vlostvolts
Example 4b — finding the internal resistance.A battery of emf 12 V is connected to a resistor. A high-resistance voltmeter across the battery reads 11 V while a current of 2 A flows. Then r = (VB − V) / I = (12 − 11) / 2 = 0.5 Ω, and the external resistance is R = V / I = 11 / 2 = 5.5 Ω.batteryconnectedresistorhigh
Example 4c — emf and r from two readings.A cell gives a current 1.0 A when connected to a 4 Ω resistor and 0.5 A when connected to a 10 Ω resistor. Find r and VB.cellgivescurrentconnected
•VB = 1.0 (4 + r) = 0.5 (10 + r)
•
  • So 4 + r = 5 + 0.5 r
  • giving 0.5 r = 1
  • r = 2 Ω
giving
•Then VB = 1.0 × (4 + 2) = 6 VThen
•Check with the second reading: 0.5 × 12 = 6 V ✔Checksecondreading
Example 4d — short-circuit current.A 1.5 V cell has an internal resistance 0.25 Ω. If its terminals are joined by a thick wire (R ≈ 0) the current is 1.5 / 0.25 = 6 A. Compare this with the 0.3 A it delivers to a 4.75 Ω lamp: 1.5 / (4.75 + 0.25) = 0.3 Acellinternalresistanceterminals
Example 4e — effect of changing R on the voltmeter.
  • A battery has emf 10 V and r = 1 Ω. When R = 4 Ω: I = 2 A
  • V = 8 V. When R = 9 Ω: I = 1 A
  • V = 9 V. When the circuit is open: V = 10 V. The reading rises towards the emf as the external resistance grows
batterycircuitopenreading
Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits · 8 / 9
Part 9

Part 9

  • 📖

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Physics Thanaweya Amma (2026/2027)

  • Lesson 2 – Resistor Combinations and Ohms Law for Closed Circuits
  • Resistors can be connected in series (to increase equivalent resistance with constant current) or parallel (to decrease equivalent resistance with constant voltage).
  • Ohm's Law for Closed Circuits calculates the terminal voltage of a non-ideal battery (V = VB – I·r).

🌳 Hierarchy Classification & Concept Tree

Resistor Networks and Closed Circuits

├── 1. Series Resistor Combination
│ ├── Constant currentItotal = I1 = I2 = I3
│ ├── Divided voltageVtotal = V1 + V2 + V3
│ └── Equivalent resistanceReq = R1 + R2 + R3
├── 2. Parallel Resistor Combination
│ ├── Constant voltageVtotal = V1 = V2 = V3
│ ├── Divided currentItotal = I1 + I2 + I3
│ └── Equivalent resistance1/Req = 1/R1 + 1/R2 + 1/R3 (Req is smaller than smallest resistor)

└── 3. Ohm's Law for Closed Circuits

├── I = VB / (Rexternal + rinternal)
├── Battery Terminal Voltage (Discharging)V = VB – I·r
└── Battery Terminal Voltage (Charging)V = VB + I·r

📋 Master Teacher's Blackboard & Exam Secrets

Comparison aspectMain ideaKeywords
RULE 1
  • When a battery is discharging
  • its terminal voltage V is LESS than its EMF (VB) due to internal voltage drop across internal resistance (I·r)
batterydischargingterminalvoltage
RULE 2
  • Terminal voltage V equals VB if: (1) The circuit is open (I = 0)
  • or (2) The internal resistance is negligible (r = 0)
Terminalvoltageequalscircuit
RULE 3
  • If external resistance R decreases
  • current I increases
  • and terminal voltage V = VB – I·r DECREASES
externalresistancedecreasescurrent
RULE 4
  • If a battery is being recharged by an external DC supply
  • its terminal voltage is GREATER than its EMF: V = VB + I·r
batterybeingrechargedexternal
📊 Comparative AnalysisComparative Synthesis: Series vs. Parallel Resistor NetworksSynthesisSeriesParallelResistor
Feature Series Combination Parallel Combination
Main Purpose To obtain a large equivalent resistance To obtain a small equivalent resistance
Current Intensity Same constant current flows through all Current branches inversely with resistance
Potential Difference Voltage divides directly with resistance Potential difference is identical across all
Equivalent Resistance Req = Σ Ri (larger than largest) 1/Req = Σ 1/Ri (smaller than smallest)
Failure of one bulb Entire circuit shuts off completely Other branches continue operating independently
Lesson 2: Resistor Combinations and Ohms Law for Closed Circuits · 9 / 9