Apply Current & Circuits — Apply B.5 Current and circuits in problem-solving and explanations - IB Physics

SECTION 1

Historical Context & Motivation

The study of electric circuits is rooted in centuries of curiosity about the nature of electricity. Early experiments with static charges and crude batteries eventually gave way to a systematic understanding of how charges move through conductors. The concepts you will use in IB Physics B.5 — current, potential difference, and resistance — were forged by a series of landmark discoveries that turned electricity from a parlor trick into the backbone of modern technology.

1800 • Volta's Pile

Alessandro Volta constructs the first true battery by stacking zinc and copper discs separated by brine-soaked cardboard, providing a steady source of continuous current for the first time.

1827 • Ohm's Law Published

Georg Simon Ohm quantifies the relationship between voltage, current, and resistance, establishing V = IR — one of the most used equations in physics.

1845 • Kirchhoff's Circuit Laws

Gustav Kirchhoff formulates his junction rule (conservation of charge) and loop rule (conservation of energy), giving physicists a complete toolkit for analyzing complex circuits.

1897 • Discovery of the Electron

J.J. Thomson identifies the electron, revealing that electric current is the flow of charged particles through a conductor, linking circuits to the particulate nature of matter.

The central question driving circuit analysis is deceptively simple: when you connect components in a loop with a battery, how much current flows, how is energy distributed, and how do series and parallel arrangements change the behavior? Answering this question requires combining the ideas of Ohm, Kirchhoff, and the microscopic picture of electron drift into a coherent problem-solving framework — exactly what IB Physics B.5 asks you to do.

SECTION 2

Core Principles & Definitions

Before solving circuit problems, you need a firm grasp of the quantities involved and the rules that govern them. The four foundational ideas below form the conceptual skeleton of every circuit question you will face in IB Physics.

Electric Current (I)

Current is the rate of flow of electric charge past a point, measured in amperes (A). Mathematically, I = ΔQ / Δt. In metals, it is free electrons that drift through the lattice in a direction opposite to conventional current.

Potential Difference (V)

Potential difference (voltage) is the energy transferred per unit charge between two points, measured in volts (V). A battery supplies an electromotive force (emf) that drives charges around the circuit.

Resistance (R)

Resistance quantifies how much a component opposes the flow of current, measured in ohms (Ω). It depends on a material's resistivity, length, and cross-sectional area: R = ρL / A.

Ohm's Law

For an ohmic conductor (constant temperature), the potential difference across it is directly proportional to the current through it: V = IR. This simple relationship is the starting point for almost every circuit calculation.

Power in Circuits

Electrical power is the rate at which energy is transferred. P = IV, and combining with Ohm's law gives P = I²R and P = V² / R. These forms let you calculate energy dissipation in any resistor.

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KEY TAKEAWAY

Think of a circuit like a water park. The pump (battery) lifts water (charge) to the top of a slide, giving it gravitational potential energy — that's like emf. As water flows down the slide (wire), friction (resistance) converts that energy into heat. The flow rate of water is like current, and the height difference is like potential difference. A narrower slide means more friction — higher resistance, less flow for the same height.

SECTION 3

Visual Explanation — Series & Parallel Circuits

Understanding how components are arranged in a circuit — in series or in parallel — is essential for IB circuit problems. The diagram below shows both arrangements side by side so you can compare how current and voltage behave in each.

Left: In a series circuit, current is identical through every component, and voltages add up to the emf. Right: In a parallel circuit, voltage is the same across each branch, and currents add up to the total.

Notice the key contrast. In the series arrangement on the left, the same current I flows through R₁ and R₂ because there is only one path. The total resistance increases, so for a given emf the current is reduced. In the parallel arrangement on the right, both resistors share the same potential difference, but the current splits — more current flows through the smaller resistor. The total resistance decreases because you have opened additional pathways for charge to flow. This behavior underpins almost every IB circuit question, so keep the diagram in mind as a mental reference.

SECTION 4

Mathematical Framework

Circuit analysis in IB Physics B.5 relies on a compact set of equations. Mastering when and how to apply each one is the difference between a quick, confident answer and a frustrating dead end.

OHM'S LAW

V = I × R

V = potential difference (V), I = current (A), R = resistance (Ω). Valid for ohmic conductors at constant temperature.

SERIES RESISTANCE

R_total = R₁ + R₂ + R₃ + …

In series, resistances simply add. The total resistance is always greater than the largest individual resistor.

PARALLEL RESISTANCE

1 / R_total = 1 / R₁ + 1 / R₂ + 1 / R₃ + …

In parallel, reciprocals add. The total resistance is always less than the smallest individual resistor.

ELECTRICAL POWER

P = I × V = I² × R = V² / R

P = power (W). Choose the form that matches the known quantities. P = I²R is handy when you know current; P = V²/R when you know voltage.

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Kirchhoff's Rules — The Complete Toolkit

Junction rule (KCL): The sum of currents entering a junction equals the sum leaving it (conservation of charge). Loop rule (KVL): Around any closed loop, the sum of all potential differences is zero (conservation of energy). These two rules let you set up simultaneous equations for circuits that are too complex for simple series/parallel reduction.

EMF AND INTERNAL RESISTANCE

ε = I × (R + r)

ε = emf of the battery (V), R = external (load) resistance, r = internal resistance of the battery. The terminal voltage across the battery is V = ε − Ir, which is less than emf whenever current flows.

SECTION 5

Kirchhoff's Laws in Action

While Ohm's law handles single-loop or easily reducible circuits, many IB problems feature multiple loops or branches that require Kirchhoff's laws. The diagram below shows a two-loop circuit with labeled currents and voltage drops, illustrating how to apply both the junction rule and the loop rule systematically.

A two-loop circuit with batteries ε₁ = 12 V and ε₂ = 6 V. The junction rule at point J gives I₁ = I₂ + I₃. Applying the loop rule to each loop yields two more equations, forming a solvable system.

The strategy for any Kirchhoff problem follows a reliable recipe. First, label all unknown currents with assumed directions — if your answer comes out negative, the actual direction is simply reversed. Second, identify every junction and write a junction-rule equation for each independent junction. Third, trace around each independent loop, summing emfs (positive when traversed from − to +) and resistor drops (−IR in the direction of assumed current). Finally, solve the resulting simultaneous equations. In IB Physics, circuits rarely have more than two loops, so you will typically end up with two or three equations — entirely manageable with substitution or elimination.

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IB Exam Tip

Always show your sign convention clearly. IB examiners award marks for correct loop equations even if the final numerical answer contains an arithmetic slip. Draw arrows for current direction on your diagram and state which direction you are traversing each loop.

SECTION 6

Worked Example — Series-Parallel Combination

A battery of emf 12 V with negligible internal resistance is connected to a circuit where a 6 Ω resistor is in series with a parallel combination of a 4 Ω and a 12 Ω resistor. Find the total resistance, the total current, and the power dissipated in the 4 Ω resistor.

Series-Parallel Circuit Analysis

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Step 1 — Identify the parallel combination

The 4 Ω and 12 Ω resistors are in parallel. Use the reciprocal formula: 1/R_p = 1/4 + 1/12 = 3/12 + 1/12 = 4/12.

R_p = 12/4 = 3 Ω
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Step 2 — Find total resistance

The 6 Ω resistor is in series with the parallel combination: R_total = 6 + 3.

R_total = 9 Ω
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Step 3 — Calculate total current

Apply Ohm's law with the battery emf: I = V / R_total = 12 / 9.

I = 1.33 A (4/3 A)
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Step 4 — Find voltage across the parallel combination

Voltage drop across the 6 Ω series resistor: V₆ = I × 6 = 1.33 × 6 = 8.0 V. The remaining voltage appears across the parallel pair: V_p = 12 − 8.0.

V_p = 4.0 V
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Step 5 — Calculate power dissipated in the 4 Ω resistor

Both parallel resistors share V_p = 4.0 V. Use P = V² / R for the 4 Ω resistor: P = (4.0)² / 4 = 16 / 4.

P = 4.0 W

SECTION 7

Series vs. Parallel — Strengths & Limitations

Choosing between series and parallel arrangements is not just a theoretical exercise — it has real engineering consequences. The table below summarizes the key differences that IB examiners test and that engineers consider when designing real circuits.

Feature	Series	Parallel
Current	Same through all components	Splits among branches; total = sum of branch currents
Voltage	Divides among components; sum = emf	Same across each branch
Total R	R increases (R₁ + R₂ + …)	R decreases (always less than smallest branch)
Component failure	One break stops all current (e.g., old holiday lights)	Other branches keep working (e.g., household wiring)
Common use	Voltage dividers, simple sensor circuits	Household power, multi-device charging

Comparison of series and parallel circuit properties

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KEY TAKEAWAY

Think of series resistors as single-lane toll booths on a highway — every car (charge) must pass through each booth, slowing traffic more with each addition. Parallel resistors are like opening extra lanes — more total traffic can flow, and if one lane closes, the others stay open. Your house uses parallel wiring so that turning off a lamp doesn't kill your refrigerator.

SECTION 8

Connection to Advanced Theory — Internal Resistance & Real Batteries

In many IB questions, you can assume an ideal battery with no internal resistance. However, real batteries have a small internal resistance r that causes the terminal voltage to drop below the emf when current flows. This concept bridges basic circuit analysis and more advanced topics such as power transfer efficiency, which appears in IB HL problems and university-level electronics.

Concept	Basic B.5 Approach	Advanced Extension
Battery model	Ideal: V = emf for all currents	Real: V = ε − Ir; terminal voltage varies with load
Power to load	P = V²/R using full emf	Maximum power transfer when R = r (matching theorem)
Energy loss	All energy goes to external resistors	Some energy is wasted as heat inside the battery (P = I²r)
Measurement	Ammeter and voltmeter give ideal readings	V vs. I graph: y-intercept = ε, gradient = −r

Basic vs. advanced battery models

As you progress beyond B.5, you will encounter AC circuits with capacitors and inductors, where impedance replaces resistance and phase angles become important. For now, the DC skills you build here — applying Ohm's law, simplifying series-parallel networks, and setting up Kirchhoff equations — form the essential foundation that every more advanced treatment relies on.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Two identical resistors are first connected in series and then in parallel across the same battery. In which configuration is the total current drawn from the battery greater, and why?

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The total current is greater in the parallel configuration. In series, R_total = 2R, giving I = V/(2R). In parallel, R_total = R/2, giving I = V/(R/2) = 2V/R. The parallel arrangement has four times as much current because the total resistance is four times smaller. More pathways mean less opposition to charge flow.

Question 2: BASIC CALCULATION

A 9.0 V battery is connected across a 15 Ω resistor. Calculate the current through the resistor and the power dissipated.

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Using Ohm's law, I = V/R = 9.0/15 = 0.60 A. Power can be found from P = IV = 0.60 × 9.0 = 5.4 W, or equivalently P = V²/R = 81/15 = 5.4 W. The resistor converts 5.4 joules of electrical energy into thermal energy every second.

Question 3: INTERMEDIATE

Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in parallel across a 12 V supply. Calculate the total resistance, the total current from the supply, and the current through the 3 Ω resistor.

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1/R_total = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so R_total = 1 Ω. Total current I = V/R_total = 12/1 = 12 A. Since all branches share 12 V, the current through the 3 Ω resistor is I₃ = 12/3 = 4.0 A. You can verify: I₂ = 12/2 = 6 A and I₆ = 12/6 = 2 A; the branch currents sum to 6 + 4 + 2 = 12 A, confirming the junction rule.

Question 4: APPLIED

A battery of emf 6.0 V and internal resistance 0.50 Ω is connected to an external resistor R. The terminal voltage measured across the battery is 5.0 V. Find R and the power dissipated inside the battery.

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The voltage drop across internal resistance is ε − V = 6.0 − 5.0 = 1.0 V. Since V_r = Ir, the current is I = 1.0/0.50 = 2.0 A. Now R = V/I = 5.0/2.0 = 2.5 Ω. The power wasted inside the battery is P_internal = I²r = (2.0)² × 0.50 = 2.0 W. This shows that 2.0 W of the battery's 12 W total output is lost to internal heating, a significant inefficiency.

Question 5: CRITICAL THINKING

A student has three 12 Ω resistors and a 9.0 V battery. She wants to design a circuit that draws exactly 1.0 A from the battery. Determine the arrangement she should use and explain your reasoning using the series and parallel resistance formulas.

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She needs R_total = V/I = 9.0/1.0 = 9.0 Ω. All three in series gives 36 Ω (too high). All three in parallel gives 4 Ω (too low). Two in parallel gives 1/R = 1/12 + 1/12 = 2/12, so R = 6 Ω. Placing this combination in series with the third 12 Ω resistor yields 6 + 12 = 18 Ω (still too high). Instead, try two in series (24 Ω) in parallel with the third (12 Ω): 1/R = 1/24 + 1/12 = 1/24 + 2/24 = 3/24 = 1/8, giving R = 8 Ω (not 9). The solution requires placing one resistor in series with a parallel pair: R = 12 + (12 × 12)/(12 + 12) = 12 + 6 = 18 Ω... this also fails. In fact, no combination of three identical 12 Ω resistors yields exactly 9 Ω. The student should recognize this and pick the closest: two in parallel (6 Ω) in series with one (12 Ω) = 18 Ω → I = 0.50 A, or all in parallel = 4 Ω → I = 2.25 A. The exact target of 1.0 A is not achievable with only three identical 12 Ω resistors, illustrating that practical circuit design sometimes requires components of different values.

SUMMARY

Lesson Summary

Summary

IB Physics B.5 centres on analysing circuits using a small but powerful toolkit. Ohm's law (V = IR) connects the three fundamental quantities — current, potential difference, and resistance. In series circuits, current stays constant while voltages add; in parallel circuits, voltage stays constant while currents add. The formulas R_total = R₁ + R₂ (series) and 1/R_total = 1/R₁ + 1/R₂ (parallel) let you simplify networks step by step.

For complex circuits, Kirchhoff's junction rule (conservation of charge) and loop rule (conservation of energy) provide a systematic way to generate equations. Real batteries introduce internal resistance that reduces terminal voltage below emf. Finally, power (P = IV = I²R = V²/R) tells you how quickly energy is transferred in each component. Master these tools and you can tackle any DC circuit problem the IB exam presents.