🎯⭐ INTERACTIVE LESSON

Current, Resistance, and Ohm's Law

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Current, Resistance, and Ohm's Law - Complete Interactive Lesson

Part 1: Electric Current

⚡ Electric Current

Part 1 of 7 — Charges in Motion

So far in electrostatics, charges have been sitting still. Now we let them move — and that moving charge is called electric current.

In this part you'll learn:

  • What electric current is and how it's measured
  • The difference between conventional current and electron flow
  • What drift velocity means (and why it's shockingly slow)
  • How to calculate current from charge and time

What Is Electric Current?

Electric current is the rate at which electric charge flows past a point in a circuit.

I=ΔQΔtI = \frac{\Delta Q}{\Delta t}

SymbolMeaningSI Unit
IICurrentAmpere (A)
ΔQ\Delta QChargeCoulomb (C)
Δt\Delta tTimeSecond (s)

The Ampere

1 A=1 C/s1 \text{ A} = 1 \text{ C/s}

One ampere means one coulomb of charge passes a point every second.

Charge Carriers

In metals, the charge carriers are free electrons (conduction electrons). Each carries charge e=1.6×1019e = 1.6 \times 10^{-19} C.

In electrolytes (salt water, batteries), both positive and negative ions can carry current.

In semiconductors, both electrons and "holes" (missing electrons) carry current.

Conventional Current vs. Electron Flow

The Historical Convention

Benjamin Franklin guessed (incorrectly) that positive charges flow through wires. We still use his convention:

Conventional current flows from high potential (+) to low potential (−).

The Reality

In a metal wire, electrons actually flow from − to + (opposite to conventional current).

Why Keep the Convention?

  • All circuit equations work perfectly with conventional current
  • The math doesn't care which sign you pick — as long as you're consistent
  • AP Physics uses conventional current unless stated otherwise

Drift Velocity

When a voltage is applied, electrons don't race through the wire. They drift slowly, bumping into atoms along the way.

The drift velocity vdv_d is typically 104\sim 10^{-4} m/s — about 0.1 mm/s!

So why does a light turn on instantly? The electric field propagates at nearly the speed of light. Every electron in the wire starts moving almost simultaneously.

Current in Terms of Drift Velocity

I=nAvdeI = nAv_d e

SymbolMeaning
nnNumber density of charge carriers (electrons/m³)
AACross-sectional area of the wire
vdv_dDrift velocity
eeCharge per carrier (1.6×10191.6 \times 10^{-19} C)

For copper: n8.5×1028n \approx 8.5 \times 10^{28} electrons/m³

Current Concepts Quiz

Current Calculation Drill

  1. A phone charger delivers 2.0 A for 1 hour. How many coulombs of charge are transferred? (in C)

  2. A lightning bolt transfers 5.0 C of charge in 2.0×1032.0 \times 10^{-3} s. What is the average current? (in A)

  3. A copper wire (cross-section 1.0×1061.0 \times 10^{-6} m², n=8.5×1028n = 8.5 \times 10^{28} m⁻³) carries 2.0 A. What is the drift velocity? (in m/s, use scientific notation like 1.5e-4)

Round all answers to 3 significant figures.

Exit Quiz

Part 2: Resistance & Resistivity

🔌 Resistance & Resistivity

Part 2 of 7 — Why Charges Slow Down

Current doesn't flow freely — every material resists it to some degree. Understanding resistance and resistivity lets you predict how much current a given voltage will push through any conductor.

What Is Resistance?

Resistance measures how much a material opposes the flow of electric current.

R=VIR = \frac{V}{I}

SymbolMeaningSI Unit
RRResistanceOhm (Ω\Omega)
VVVoltageVolt (V)
IICurrentAmpere (A)

1  Ω=1  V/A1 \;\Omega = 1 \;\text{V/A}

Physical Picture

Think of resistance like friction for charges. As electrons drift through a conductor, they collide with the vibrating lattice of atoms. Each collision:

  • Transfers kinetic energy to the lattice (→ heat)
  • Slows the electron down before the electric field accelerates it again

More collisions → more resistance → less current for a given voltage.

Resistivity and the Resistance Formula

Resistance depends on both the material and the geometry of the conductor:

R=ρLAR = \frac{\rho L}{A}

SymbolMeaningSI Unit
ρ\rhoResistivityΩm\Omega \cdot \text{m}
LLLengthm
AACross-sectional area

What Each Factor Does

ChangeEffect on RRWhy?
Longer wireRR increasesMore material for electrons to travel through
Thicker wireRR decreasesMore "lanes" for electrons to flow
Higher ρ\rhoRR increasesMaterial itself resists more

Typical Resistivities (at 20°C)

Materialρ\rho (Ωm\Omega \cdot \text{m})Type
Silver1.59×1081.59 \times 10^{-8}Conductor
Copper1.68×1081.68 \times 10^{-8}Conductor
Aluminum2.65×1082.65 \times 10^{-8}Conductor
Nichrome1.10×1061.10 \times 10^{-6}Alloy (heating elements)
Silicon640640Semiconductor
Glass101010^{10}101410^{14}Insulator
Rubber1013\sim 10^{13}Insulator

Conductors vs. Insulators vs. Semiconductors

  • Conductors (ρ108\rho \sim 10^{-8}): many free electrons, very low resistance
  • Semiconductors (ρ101\rho \sim 10^{-1} to 10310^{3}): few free carriers at room temp; resistance decreases with temperature
  • Insulators (ρ>108\rho > 10^{8}): almost no free carriers, extremely high resistance

Resistance Concepts Quiz

Resistance Proportionality Check 🎯

Resistance Calculation Drill 🔧

Use ρCu=1.68×108\rho_{\text{Cu}} = 1.68 \times 10^{-8} Ω\Omega\cdotm.

  1. A copper wire is 10.0 m long with cross-sectional area 2.0×1062.0 \times 10^{-6} m². What is its resistance? (in Ω\Omega)

  2. A copper wire has resistance 0.50 Ω\Omega and length 5.0 m. What is its cross-sectional area? (in m², scientific notation like 1.7e-7)

  3. You need a copper wire with resistance exactly 1.0 Ω\Omega and diameter 1.0 mm. How long must it be? (in m, round to nearest whole number)

Exit Quiz

Part 3: Ohm's Law & Power

📐 Ohm's Law & Electric Power

Part 3 of 7 — The Most Important Equation in Circuits

Ohm's Law connects voltage, current, and resistance in one elegant equation. Combined with the power formulas, you can analyze any simple circuit.

Ohm's Law

V=IRV = IR

This says: the voltage drop across a resistor equals the current through it times its resistance.

Three Forms

Solving forFormula
VoltageV=IRV = IR
CurrentI=V/RI = V/R
ResistanceR=V/IR = V/I

Ohmic vs. Non-Ohmic Materials

Ohmic materials obey V=IRV = IR with constant RR:

  • Metals at constant temperature
  • Carbon resistors
  • Their II-VV graph is a straight line through the origin

Non-ohmic materials have resistance that changes:

  • Light bulbs (filament heats up → RR increases)
  • Diodes (current flows in only one direction)
  • Semiconductors
  • Their II-VV graph is curved

Reading II-VV Curves

On an II vs. VV graph:

  • Slope = 1/R1/R (for ohmic materials, the slope is constant)
  • Steeper line = lower resistance (more current for same voltage)
  • Shallower line = higher resistance

Electric Power

Power is the rate at which electrical energy is converted to other forms (heat, light, motion):

P=IVP = IV

Combining with Ohm's Law gives three equivalent forms:

FormulaBest when you know
P=IVP = IVCurrent and voltage
P=I2RP = I^2 RCurrent and resistance
P=V2/RP = V^2 / RVoltage and resistance

Units

1 W=1 VA=1 A2Ω=1 V2/Ω1 \text{ W} = 1 \text{ V} \cdot \text{A} = 1 \text{ A}^2 \cdot \Omega = 1 \text{ V}^2 / \Omega

Key Insight

In a resistor, electrical energy is converted entirely to thermal energy (heat). This is called Joule heating or resistive dissipation.

A 100 W light bulb converts 100 joules of electrical energy per second — mostly to heat, with a small fraction as visible light.

Ohm's Law Concept Check

Power Concept Check

Ohm's Law & Power Drill 🔧

  1. A 100 Ω\Omega resistor carries 0.30 A. What is the voltage across it? (in V)

  2. A toaster draws 10 A from a 120 V outlet. What is its resistance? (in Ω\Omega)

  3. What power does the toaster in #2 dissipate? (in W)

  4. A 500 Ω\Omega resistor has 25 V across it. What power does it dissipate? (in W)

Round all answers to 3 significant figures.

Exit Quiz

Part 4: Temperature Dependence

🌡️ Resistivity & Temperature

Part 4 of 7 — Why Hot Wires Resist More

Resistance isn't fixed — it changes with temperature. Understanding this relationship is essential for designing circuits that work reliably and for understanding exotic phenomena like superconductivity.

Temperature Dependence of Resistivity

For most metals, resistivity increases approximately linearly with temperature:

ρ=ρ0(1+αΔT)\rho = \rho_0 \bigl(1 + \alpha \Delta T\bigr)

SymbolMeaning
ρ\rhoResistivity at temperature TT
ρ0\rho_0Resistivity at reference temperature T0T_0 (usually 20°C)
α\alphaTemperature coefficient of resistivity (°C⁻¹)
ΔT\Delta TTT0T - T_0 (temperature change)

Since R=ρL/AR = \rho L/A and the geometry changes are usually negligible:

R=R0(1+αΔT)R = R_0 \bigl(1 + \alpha \Delta T\bigr)

Typical Temperature Coefficients

Materialα\alpha (°C⁻¹)
Silver3.8×1033.8 \times 10^{-3}
Copper3.9×1033.9 \times 10^{-3}
Aluminum3.9×1033.9 \times 10^{-3}
Tungsten4.5×1034.5 \times 10^{-3}
Nichrome0.4×1030.4 \times 10^{-3}
Carbon0.5×103-0.5 \times 10^{-3}
Silicon75×103-75 \times 10^{-3}

Why Metals Have Positive α\alpha

Higher temperature → atoms vibrate more → more collisions with drifting electrons → higher resistivity.

Why Semiconductors Have Negative α\alpha

Higher temperature → more electrons gain enough energy to become free carriers → more charge carriers → lower resistivity (despite more collisions).

Superconductors

At very low temperatures, some materials have their resistance drop to exactly zero.

Key Facts

  • Below a critical temperature TcT_c, resistance = 0
  • Current flows indefinitely with no energy loss
  • Mercury: Tc=4.2T_c = 4.2 K (discovered 1911)
  • High-temperature superconductors: Tc90T_c \sim 90130130 K (still very cold!)
  • Applications: MRI magnets, particle accelerators, maglev trains

Practical Wire Sizing (AWG)

American Wire Gauge (AWG) is the standard system for wire sizes in the US:

AWGDiameter (mm)Typical Use
141.6315 A circuits (lighting)
122.0520 A circuits (outlets)
102.5930 A circuits (dryers)
64.1160 A circuits (ranges)

Smaller AWG number = thicker wire = lower resistance = higher current capacity.

Why does wire gauge matter? If a wire is too thin for the current it carries, P=I2RP = I^2 R causes excessive heating — a fire hazard!

Temperature & Resistance Quiz

Temperature-Resistance Drill 🌡️

  1. A copper wire (α=3.9×103\alpha = 3.9 \times 10^{-3} °C⁻¹) has resistance 5.0 Ω\Omega at 20°C. What is its resistance at 120°C? (in Ω\Omega)

  2. An aluminum wire (α=3.9×103\alpha = 3.9 \times 10^{-3} °C⁻¹) has resistance 10.0 Ω\Omega at 20°C. At what temperature will its resistance be 15.0 Ω\Omega? (in °C)

  3. A carbon resistor (α=0.5×103\alpha = -0.5 \times 10^{-3} °C⁻¹) has resistance 1000 Ω\Omega at 20°C. What is its resistance at 220°C? (in Ω\Omega)

Round all answers to 3 significant figures.

Exit Quiz

Part 5: Electric Power & Energy

💡 Electric Power & Energy

Part 5 of 7 — Paying for Electrons

Every electrical device converts energy from one form to another. Understanding power and energy lets you calculate how much energy a device uses — and how much it costs to run.

Power Formulas (Review & Extension)

Recall the three forms of the power equation:

P=IV=I2R=V2RP = IV = I^2R = \frac{V^2}{R}

Energy

Power is the rate of energy use, so:

E=PtE = Pt

SymbolMeaningSI Unit
EEEnergyJoule (J)
PPPowerWatt (W)
ttTimeSecond (s)

The Kilowatt-Hour

Your electric company doesn't bill you in joules — they use kilowatt-hours (kWh):

1 kWh=1000 W×3600 s=3.6×106 J=3.6 MJ1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J} = 3.6 \text{ MJ}

A kilowatt-hour is the energy used by a 1000 W device running for 1 hour.

Typical Electricity Cost

In the US: approximately 0.120.12–0.15 per kWh (varies by region).

Cost=Energy (kWh)×Rate ($/kWh)\text{Cost} = \text{Energy (kWh)} \times \text{Rate (\$/kWh)}

Household Circuits

Standard US Household Power

PropertyValue
Voltage120 V (standard) or 240 V (large appliances)
Frequency60 Hz (AC)
Typical circuit breaker15 A or 20 A

Maximum Power per Circuit

For a 15 A, 120 V circuit:

Pmax=IV=(15)(120)=1800 WP_{\text{max}} = IV = (15)(120) = 1800 \text{ W}

This is why you can't run too many high-power devices on one circuit!

Common Appliance Power Ratings

AppliancePower (W)
LED bulb10
Laptop50–100
Hair dryer1200–1800
Microwave1000–1200
Space heater1500
Electric oven2000–5000 (240 V circuit)
Central AC3000–5000 (240 V circuit)

Why Large Appliances Use 240 V

For the same power: P=IVP = IV, doubling VV halves II. Lower current means:

  • Thinner (cheaper) wires
  • Less I2RI^2R heating loss in the wires
  • Smaller circuit breakers

Power & Energy Quiz

Energy & Cost Drill 💰

Use an electricity rate of $0.12/kWh.

  1. A 100 W light bulb runs 24 hours a day for 30 days. How many kWh does it use?

  2. What is the monthly cost for that light bulb? (in $, round to nearest cent)

  3. A 5000 W electric dryer runs for 45 minutes. How many kWh does it use?

  4. You replace ten 60 W incandescent bulbs with ten 9 W LED bulbs. How many kWh do you save per day if they run 6 hours/day?

  5. At 0.12/kWh,howmuchdoyousaveperyear(365days)fromthatLEDswap?(in0.12/kWh, how much do you save per year (365 days) from that LED swap? (in , round to nearest dollar)

Round all answers to 3 significant figures.

Exit Quiz

Part 6: EMF & Internal Resistance

🔋 Real-World Applications

Part 6 of 7 — Batteries, Bulbs, and Safety

Real circuits aren't ideal. Batteries have internal resistance, wires have finite conductivity, and too much current can be dangerous. Let's see how the theory connects to the real world.

Batteries: EMF & Internal Resistance

A real battery isn't a perfect voltage source. It has:

  • EMF (ε\varepsilon): the "ideal" voltage the battery would supply with no current flowing (open-circuit voltage)
  • Internal resistance (rr): resistance inside the battery itself

Terminal Voltage

When current II flows through the battery:

Vterminal=εIrV_{\text{terminal}} = \varepsilon - Ir

The internal resistance causes a voltage drop inside the battery, so the voltage available to the external circuit is less than the EMF.

Key Relationships

ConditionTerminal Voltage
No current (I=0I = 0)V=εV = \varepsilon (full EMF)
Small currentVεV \approx \varepsilon (small drop)
Large currentVεV \ll \varepsilon (significant drop)
Short circuit (Rext=0R_{\text{ext}} = 0)I=ε/rI = \varepsilon / r (maximum current, dangerous!)

Current in the Full Circuit

With external resistance RR:

I=εR+rI = \frac{\varepsilon}{R + r}

The total resistance is always R+rR + r, and the EMF drives the current through both.

Power Budget

Ptotal=εI=I2R+I2rP_{\text{total}} = \varepsilon I = I^2R + I^2r

  • I2RI^2R: useful power delivered to external circuit
  • I2rI^2r: wasted as heat inside the battery

Circuit Safety Devices

Fuses

A fuse contains a thin wire that melts when current exceeds its rating, breaking the circuit.

  • One-time use — must be replaced after blowing
  • Rated by current (e.g., 15 A, 20 A)
  • The thin wire has high resistance → heats up quickly at high current

Circuit Breakers

A circuit breaker uses a bimetallic strip or electromagnet to trip (open) the circuit when current is too high.

  • Resettable — just flip the switch
  • Standard in modern homes
  • Also rated by current

Ground Fault Circuit Interrupter (GFCI)

A GFCI detects when current flowing out differs from current flowing back (meaning some current is leaking through a person or water).

  • Trips in milliseconds
  • Required near water: bathrooms, kitchens, outdoors
  • Can detect leakage as small as 5 mA

Why These Matter

Current through bodyEffect
1 mABarely perceptible
10 mAPainful shock
100 mACan be fatal (ventricular fibrillation)
1 ASevere burns, likely fatal

Safety devices prevent dangerous currents from flowing through the circuit — and through you.

Battery & EMF Quiz

Safety Concepts Quiz

Real-World Application Check 🎯

Exit Quiz

Part 7: Synthesis & AP Review

🎯 Synthesis & AP Review

Part 7 of 7 — Putting It All Together

You've learned about current, resistance, resistivity, Ohm's Law, power, and real-world applications. Now let's connect everything and prepare for the AP exam.

Concept Map: Current, Resistance & Ohm's Law

The Core Equations

EquationWhat It Describes
I=ΔQ/ΔtI = \Delta Q / \Delta tCurrent = charge flow rate
I=nAvdeI = nAv_d eCurrent from drift velocity
R=ρL/AR = \rho L / AResistance from geometry & material
ρ=ρ0(1+αΔT)\rho = \rho_0(1 + \alpha \Delta T)Resistivity vs. temperature
V=IRV = IROhm's Law
P=IV=I2R=V2/RP = IV = I^2R = V^2/RElectrical power
E=PtE = PtEnergy consumed
V=εIrV = \varepsilon - IrBattery terminal voltage

How They Connect

  1. A battery (ε\varepsilon) drives current through a circuit
  2. Current (II) depends on total resistance: I=ε/(R+r)I = \varepsilon / (R + r)
  3. Resistance (RR) depends on material (ρ\rho), length (LL), area (AA), and temperature
  4. Power dissipated in each element: P=I2RP = I^2R
  5. Energy over time: E=PtE = Pt

Common Mistakes to Avoid

MistakeCorrection
Forgetting internal resistanceTotal RR = Rext+rR_{\text{ext}} + r
Using wrong power formulaMatch formula to known quantities
Confusing ρ\rho (resistivity) with RR (resistance)ρ\rho is a material property; RR depends on geometry too
Thinking drift velocity = signal speedDrift velocity is 104\sim 10^{-4} m/s; signal propagates at c\sim c
Mixing up conventional current and electron flowConventional: + to −. Electrons: − to +
Forgetting units on α\alphaα\alpha has units of °C⁻¹ (or K⁻¹)

Mixed Concept Check

Mixed Problem Drill 🧮

  1. A nichrome wire (ρ=1.10×106  Ω\rho = 1.10 \times 10^{-6}\;\Omega\cdotm) is 2.0 m long with diameter 0.50 mm. What is its resistance? (in Ω\Omega, round to 1 decimal)

  2. A battery (ε=12\varepsilon = 12 V, r=0.40  Ωr = 0.40\;\Omega) is connected to a 5.6  Ω5.6\;\Omega resistor. What current flows? (in A)

  3. What is the terminal voltage of the battery in #2? (in V)

  4. A 1200 W hair dryer runs on 120 V for 15 minutes. How much energy does it use? (in kJ)

Round all answers to 3 significant figures.

AP FRQ Preview

On the AP Physics 2 exam, you'll encounter free-response questions that combine multiple concepts. Here's the type of reasoning you'll need:

Example FRQ Scenario

A student has a battery of unknown EMF and internal resistance. She connects it to a variable external resistor and measures both the terminal voltage and the current for several resistance values.

Part (a): Explain how to determine ε\varepsilon and rr from a graph of VV vs. II.

Key insight: V=εIrV = \varepsilon - Ir is a linear equation of the form y=b+mxy = b + mx:

  • y-intercept (I=0I = 0): V=εV = \varepsilon → gives EMF
  • Slope: r-r → gives internal resistance
  • x-intercept (V=0V = 0): I=ε/rI = \varepsilon/r → gives short-circuit current

Part (b): The student wants to maximize the power delivered to the external resistor. What value of RR should she use?

Key insight: PR=I2R=[ε/(R+r)]2RP_R = I^2 R = [\varepsilon/(R+r)]^2 R. Taking the derivative and setting it to zero gives R=rR = r (maximum power transfer theorem).

Part (c): Why is the "efficiency" (fraction of power delivered externally) only 50% at maximum power transfer?

Key insight: When R=rR = r, the current is I=ε/2rI = \varepsilon/2r. Power to load: I2R=ε2/4rI^2R = \varepsilon^2/4r. Total power: εI=ε2/2r\varepsilon I = \varepsilon^2/2r. Efficiency = 50%.

AP Exam Tips

  1. Show your work — write the equation, substitute, solve
  2. Include units in every answer
  3. Justify qualitative answers with equations
  4. Sketch graphs when asked — label axes and key features
  5. Check limiting cases — does your answer make sense when R0R \to 0 or RR \to \infty?

Final Mastery Quiz 🏆