Electricity & Magnetism — Science

  1. Grade 8 — Electricity & Simple Circuits
  2. Grade 12 — Electric Charge & Fields
  3. Grade 12 — Current, Voltage & Resistance
  4. Grade 12 — Circuits & Ohm's Law
  5. Grade 12 — Magnetism & Electromagnetism

Apply it in a new setting

Problem 1

A small appliance draws a current of 0.4 A when plugged into a 120 V outlet. What is its resistance? Use Ohm's law V = IR.
Hints

Rearrange V = IR to solve for resistance R.

R = V ÷ I.

R = 120 ÷ 0.4.

Answer and explanation

300 Ω

From V = IR, resistance R = V ÷ I.

R = 120 ÷ 0.4.

R = 300 Ω.

Problem 2

A hair dryer draws 10 A of current when connected to a 120 V supply. What is its power consumption? Use P = IV.
Hints

Electrical power is current multiplied by voltage.

P = I · V.

P = 10 · 120.

Answer and explanation

1200 W

Power P = IV.

P = 10 · 120.

P = 1200 W.

Problem 3

A light bulb rated 60 W runs on a 120 V circuit. What is the resistance of its filament? First find the current with P = IV, then use V = IR.
Hints

Start by finding the current: from P = IV, I = P ÷ V.

I = 60 ÷ 120 = 0.5 A.

Then R = V ÷ I = 120 ÷ 0.5.

Answer and explanation

240 Ω

Current I = P ÷ V = 60 ÷ 120 = 0.5 A.

Resistance R = V ÷ I = 120 ÷ 0.5.

R = 240 Ω.

Problem 4

Three resistors of 10 Ω, 20 Ω, and 30 Ω are connected in series across a 12 V battery. What current flows through the circuit? First find the total series resistance, then use Ohm's law.
Hints

In series, resistances simply add up.

R_total = 10 + 20 + 30 = 60 Ω.

Then I = V ÷ R_total = 12 ÷ 60.

Answer and explanation

0.2 A

Series resistances add: R_total = 10 + 20 + 30 = 60 Ω.

Ohm's law: I = V ÷ R = 12 ÷ 60.

I = 0.2 A.

Problem 5

A 6 Ω resistor and a 3 Ω resistor are connected in parallel. What is the equivalent resistance of the pair? For two resistors in parallel, R = (R_1 · R_2) ÷ (R_1 + R_2).
Hints

The parallel combination is always smaller than the smallest resistor.

R = (R_1 · R_2) ÷ (R_1 + R_2) = (6 · 3) ÷ (6 + 3).

That is 18 ÷ 9.

Answer and explanation

2 Ω

For two resistors in parallel, R = (R_1 · R_2) ÷ (R_1 + R_2).

R = (6 · 3) ÷ (6 + 3) = 18 ÷ 9.

R = 2 Ω.

Problem 6

A 1500 W space heater is plugged into a 120 V household circuit. How much current does it draw? Use P = IV. (A standard circuit breaker trips at 15 A — this draw is close to the limit.)
Hints

Rearrange P = IV to solve for the current I.

I = P ÷ V.

I = 1500 ÷ 120.

Answer and explanation

12.5 A

From P = IV, current I = P ÷ V.

I = 1500 ÷ 120.

I = 12.5 A (just under the 15 A breaker limit).

Problem 7

In an older string of holiday lights wired in series, a single bulb burns out and the entire string goes dark. Which statement best explains why?
  1. A series circuit provides only one path for current, so a single break stops the current everywhere.
  2. The remaining bulbs used up all of the available current before it reached the rest.
  3. Series circuits store extra voltage that is released and burns out every bulb at once.
  4. The broken bulb increased the current so much that all the others switched off.
Hints

Think about how many separate paths the current has in a series circuit.

If there is only one loop, what happens when the loop is broken anywhere?

Compare this to parallel wiring, where each bulb has its own path and others stay lit.

Answer and explanation

A series circuit provides only one path for current, so a single break stops the current everywhere.

A series circuit is a single loop, giving current just one path to follow.

When one bulb's filament breaks, that single path is broken, so no current flows anywhere.

Every bulb goes dark because the one path is open, so A is correct.

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