Unit 9 · Lesson 1

🔌Voltage, Current & Resistance

The Three Fundamental Quantities

Think of electric circuits like water in pipes: - Voltage (V): the "pressure" pushing charges through the circuit. Measured in Volts (V). Also called potential difference. - Current (I): the flow rate of charge — Coulombs per second. Measured in Amperes (A = C/s). - Resistance (R): opposition to current flow, like the narrowness of a pipe. Measured in Ohms (Ω).

Voltage is provided by batteries or power supplies. It's the "energy per unit charge" given to charges.

Current flows from high potential (positive terminal) to low potential (negative terminal) — called conventional current direction.

Resistance depends on material, length, and cross-sectional area: R = ρL/A

Electric Power

Electric power is the rate at which energy is delivered to a circuit element:

P = IV = I²R = V²/R

Units: Watts (W = J/s)

Your home appliances are rated in watts. A 60W light bulb uses energy at 60 J/s.

The energy used over time: E = P × t (in Joules, or kWh for electric bills)

🤔

Think About It

If you double the voltage in a circuit while keeping resistance the same, what happens to current? What happens to power?

📐 Key Equations

Fundamental circuit quantities

V = IR (Ohm's Law)
P = IV = I²R = (V²)/(R)
E = Pt

⚠️ Common Mistakes

Misconception: Voltage and current are the same thing.

✓ Correct thinking: Voltage is the "pressure" (energy per charge) that drives current; current is the actual flow of charge. They are related by resistance: V = IR.

Why: The water analogy helps: voltage is water pressure, current is flow rate. You can have high pressure with low flow (narrow pipe = high resistance).

Misconception: A battery provides constant current regardless of the circuit.

✓ Correct thinking: A battery provides a (roughly) constant voltage. The current depends on the total resistance: I = V/R.

Why: Adding more resistance to the circuit reduces the current. The battery's voltage stays the same, but the current adjusts.

Misconception: Power is always P = IV, so doubling current doubles power.

✓ Correct thinking: Doubling current while keeping resistance constant actually quadruples power, since P = I²R.

Why: When you double I with the same R, voltage also doubles (V = IR), so P = IV doubles twice — power goes up by a factor of 4.

📝 Practice Problems

Try these problems. Check your answer when ready.

#1

A 9 V battery is connected to a 3 Ω resistor. What current flows through the resistor?

easy
#2

A current of 2 A flows through a 5 Ω resistor. What is the voltage across it?

easy
#3

A 60 W light bulb is plugged into a 120 V outlet. What current does it draw?

easy
P = IV ⇒ I = P/V
#4

A resistor dissipates 18 W when a 3 A current flows through it. What is its resistance?

medium
P = I²R ⇒ R = P/I²
#5

A toaster draws 8 A from a 120 V outlet. How much energy does it use in 3 minutes?

medium
E = Pt = IVt
#6

Two devices are rated 100 W and 60 W both at 120 V. Which has more resistance? Calculate both.

hard
R = V²/P

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