Electricity & Simple Circuits — Practice · Grade 8 Science

Chapter: Energy & Heat Transfer

What this practice covers

By the end of Electricity & Simple Circuits you should be able to:

12 questions, numeric, multiple choice and written answer, each with a worked explanation.

Vocabulary used: circuit, current, voltage, resistance, series circuit, parallel circuit, model, evidence, system boundary, assumption.

Problem 1

A circuit has 9 volts and 3 ohms of resistance. Find the current.
Answer and explanation

3 A

I = V ÷ R = 9 ÷ 3 = 3 A.

Problem 2

A current of 4 amps flows through a 6-ohm resistor. Find the voltage.
Answer and explanation

24 V

V = IR = 4 × 6 = 24 V.

Problem 3

A 15-volt circuit has a current of 5 amps. Find the resistance.
Answer and explanation

3 Ω

R = V ÷ I = 15 ÷ 5 = 3 Ω.

Problem 4

In a series circuit, if one bulb burns out…
  1. only that bulb goes out; the others stay lit
  2. all the bulbs go out because the circuit is broken
  3. the other bulbs get brighter
  4. nothing changes
Answer and explanation

all the bulbs go out because the circuit is broken

A series circuit has only one path, so a break anywhere stops current everywhere on that path.

Problem 5

Which material is the BEST electrical conductor?
  1. rubber
  2. glass
  3. copper
  4. wood
Answer and explanation

copper

Copper is a metal that conducts electricity very well, which is why it's used in wiring.

Problem 6

Explain one advantage of parallel circuits over series circuits, using your home's wiring as an example.
Answer and explanation

In a parallel circuit, each device, like a lamp or TV, has its own path, so if one device is switched off or stops working, the others keep getting power. That's why turning off one light at home doesn't turn off the whole house.

Problem 7

Two bulbs are in series and one connection is removed. What happens? Compare with two independent parallel branches. Explain the reasoning in your own words.
Answer and explanation

Both series bulbs go out; a bulb in the other intact parallel branch can stay lit.

Breaking the series path stops current through both bulbs.

Opening one parallel branch can leave the other branch's complete path intact.

Response criteria

  • Use the stated conditions.
  • Show the evidence or calculation.
  • Explain the conclusion and any stated limitation.

Problem 8

A 9 V source connects across an 18 Ω resistor. Find current. Show the reasoning, not just the conclusion.
Hints

Check 0.5×12=6 V.

Answer and explanation

I=V/R=9/18=0.5 A in the ideal model.

Problem 9

Why does a gap in a simple series circuit stop steady current throughout the loop?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

The gap breaks the continuous conducting path; charge cannot circulate steadily through that loop.

Problem 10

Complete source/return diagrams, compare the three supplied records and explain evidence for energy transfer by current. Identify which observations would be needed to measure energy quantitatively.
Hints

Use the illustrated study titled Observe energy transfer in circuit paths. Keep the observations, model and conditions separate.

Answer and explanation

With the supplied series loop closed, both lamps respond; opening the single path makes both go out. In the parallel model, opening one branch leaves the other complete branch operating. Light/thermal responses provide qualitative energy-transfer evidence, not charge consumption. Quantitative energy needs voltage, current and duration or another justified method, not brightness alone.

With the supplied series loop closed, both lamps respond; opening the single path makes both go out. In the parallel model, opening one branch leaves the other complete branch operating. Light/thermal responses provide qualitative energy-transfer evidence, not charge consumption. Quantitative energy needs voltage, current and duration or another justified method, not brightness alone.

Response criteria

  • Complete conducting paths and distinguish arrangements.
  • Use observed device behavior as qualitative evidence.
  • Separate charge, energy and measurement limits.

Problem 11

Current is charge per time; voltage is energy per chargeCurrent is charge per time; voltage is energy per chargeCharge flow: 4 C in 2 s → current 2 AEnergy per charge: 12 J / 4 C → potential difference 3 VBattery energy transferLamp / receiver
This conceptual closed circuit separates a supplied charge-flow record, 4 coulombs in 2 seconds, from a supplied energy-per-charge record, 12 joules for 4 coulombs across the receiver. They give 2 A and 3 V respectively. The arrow follows conventional current rather than a drawn individual electron path. Charges circulate; a lamp transfers energy rather than using up the charge itself.
For this unit’s illustrated case, what current and potential difference follow from the two records, and why are they different quantities?
Answer and explanation

Current = 4 C/2 s = 2 A; potential difference = 12 J/4 C = 3 V. One is charge-flow rate and the other energy transferred per charge.

A coulomb (C) is a unit of electric charge; an ampere (A) is one coulomb per second, and a volt (V) is one joule per coulomb. Introduce these units before using their ratios. The diagram uses a safe supplied-data circuit, not mains wiring.

Current = 4 C/2 s = 2 A; potential difference = 12 J/4 C = 3 V. One is charge-flow rate and the other energy transferred per charge.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 12

A supplied receiver transfers 18 J for 6 C of charge while 6 C passes in 3 s. Find potential difference and current, then explain why the receiver does not use up the charge.
Answer and explanation

Potential difference 18/6 = 3 V; current 6/3 = 2 A. Charge circulates through the complete circuit while energy is transferred to the receiver.

Use joules per coulomb for voltage and coulombs per second for current.

Keep conserved charge flow distinct from energy transfer.

Potential difference 18/6 = 3 V; current 6/3 = 2 A. Charge circulates through the complete circuit while energy is transferred to the receiver.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

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