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Electricity: Current, Voltage, Resistance and Ohm's Law — Answers

Answer sheet with model answers and success criteria for the Year 9 energy, light, sound & electricity worksheet.

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Success TutoringWorksheetsAnswer sheetYear 9 · Science · Energy, Light, Sound & Electricity

Electricity: Current, Voltage, Resistance and Ohm's Law — Answers

Part A: Current, voltage and resistance

Current (I) is the rate of flow of electric charge around a circuit, measured in amperes (A, or amps). 1 A = 1000 milliamps (mA).

Voltage (V), also called potential difference, is the energy given to or used by each unit of charge, measured in volts (V). The battery provides the voltage that pushes the current.

Resistance (R) is how much a component opposes the flow of current, measured in ohms. (Its symbol is the Greek capital letter omega; this worksheet writes the unit as ohms.)

Ohm's law: for a resistor at constant temperature, V = IR. Rearranged: I = V ÷ R and R = V ÷ I.

1.Match each quantity to its unit and measuring instrument.
  • Current (I)
  • Voltage (V)
  • Resistance (R)
  • Volts (V), measured with a voltmeter
  • Ohms, calculated from V ÷ I or measured with an ohmmeter
  • Amperes (A), measured with an ammeter

Answer: Current (I) → Amperes (A), measured with an ammeter; Voltage (V) → Volts (V), measured with a voltmeter; Resistance (R) → Ohms, calculated from V ÷ I or measured with an ohmmeter

2.How must an ammeter be connected to measure the current through a lamp?
  1. AIn series with the lamp
  2. BIn parallel with the lamp
  3. CAcross the battery only
  4. DIt does not matter

Answer: In series with the lamp

The current has to flow through the ammeter, so it is placed in the same loop as the lamp (in series).

3.How must a voltmeter be connected to measure the voltage across a lamp?
  1. AIn parallel with the lamp (across it)
  2. BIn series with the lamp
  3. CAnywhere in the circuit
  4. DBetween two wires that are not connected

Answer: In parallel with the lamp (across it)

A voltmeter compares the energy of the charges on each side of a component, so it is connected across (in parallel with) that component.

4.Conventional current is shown flowing from the positive terminal of the battery, around the circuit, to the negative terminal.

Answer: True

This convention was set before electrons were discovered. The electrons in the wires actually flow the other way, from negative to positive.

Part B: Using V = IR

Write the formula, substitute with units, and give your answer with the correct unit.

5.A current of 2 A flows through a 6-ohm resistor. What is the voltage across it?

Answer: 12 V

  1. V = IR
  2. = 2 A × 6 ohms = 12 V
6.A 9 V battery is connected to a 3-ohm resistor. What current flows?

Answer: 3 A

  1. I = V ÷ R
  2. = 9 V ÷ 3 ohms = 3 A
7.A lamp has 12 V across it and a current of 0.5 A through it. What is its resistance?

Answer: 24 ohms

  1. R = V ÷ I
  2. = 12 V ÷ 0.5 A = 24 ohms
8.A resistor has 4.5 V across it and a current of 0.15 A through it. What is its resistance?

Answer: 30 ohms

  1. R = V ÷ I
  2. = 4.5 V ÷ 0.15 A = 30 ohms
9.A current of 200 mA flows through a 50-ohm resistor. What is the voltage across it?

Answer: 10 V

  1. Convert to amps: 200 mA = 200 ÷ 1000 = 0.2 A
  2. V = IR = 0.2 A × 50 ohms = 10 V
10.A 6 V supply is connected to a 1.5-ohm heating coil. What current flows?

Answer: 4 A

  1. I = V ÷ R
  2. = 6 V ÷ 1.5 ohms = 4 A

Part C: Series and parallel calculations

Series circuit (one loop): the current is the same through every component; the voltages across the components add up to the supply voltage; total resistance = sum of the resistances.

Parallel circuit (two or more branches): each branch has the full supply voltage across it; the branch currents add up to the total current from the supply; the total resistance is less than the smallest branch resistance.

11.A 12 V battery is connected in series with a 2-ohm resistor and a 4-ohm resistor. Find the total resistance, the current and the voltage across each resistor.

Answer: Total resistance 6 ohms; current 2 A; 4 V across the 2-ohm resistor and 8 V across the 4-ohm resistor

  1. Total resistance = 2 ohms + 4 ohms = 6 ohms
  2. I = V ÷ R = 12 V ÷ 6 ohms = 2 A (the same everywhere in a series circuit)
  3. V across the 2-ohm resistor = 2 A × 2 ohms = 4 V; V across the 4-ohm resistor = 2 A × 4 ohms = 8 V
  4. Check: 4 V + 8 V = 12 V, the supply voltage
12.A 9 V battery is connected in series with resistors of 10 ohms, 20 ohms and 15 ohms. Find the current and the voltage across each resistor.

Answer: Current 0.2 A; voltages 2 V, 4 V and 3 V

  1. Total resistance = 10 + 20 + 15 = 45 ohms
  2. I = 9 V ÷ 45 ohms = 0.2 A
  3. 10-ohm resistor: 0.2 × 10 = 2 V; 20-ohm resistor: 0.2 × 20 = 4 V; 15-ohm resistor: 0.2 × 15 = 3 V
  4. Check: 2 + 4 + 3 = 9 V
13.A 6-ohm resistor and a 3-ohm resistor are connected in parallel to a 12 V battery. Find the current in each branch, the total current from the battery and the total resistance of the circuit.

Answer: 6-ohm branch: 2 A; 3-ohm branch: 4 A; total current 6 A; total resistance 2 ohms

  1. In parallel, each branch has the full 12 V across it
  2. 6-ohm branch: I = 12 ÷ 6 = 2 A; 3-ohm branch: I = 12 ÷ 3 = 4 A
  3. Total current = 2 A + 4 A = 6 A
  4. Total resistance = V ÷ I = 12 V ÷ 6 A = 2 ohms (less than the smallest branch resistance)

Part D: Analysing circuits

14.A student measures the current through a resistor for different voltages. Voltage (V): 0, 1.5, 3.0, 4.5, 6.0. Current (A): 0, 0.10, 0.20, 0.30, 0.40. Find the resistance and describe the shape of a graph of voltage against current.

Answer: 15 ohms; a straight line through the origin

  1. R = V ÷ I: 1.5 ÷ 0.10 = 15 ohms; 3.0 ÷ 0.20 = 15 ohms; 4.5 ÷ 0.30 = 15 ohms; 6.0 ÷ 0.40 = 15 ohms
  2. The resistance is constant at 15 ohms, so the graph is a straight line through the origin
15.The resistance of a filament lamp stays the same as the current through it increases.

Answer: False

As the current increases, the filament gets hotter and its resistance increases, so a graph of voltage against current for a lamp is a curve, not a straight line.

16.Two identical lamps are connected in series to a battery. One lamp breaks. What happens to the other lamp?
  1. AIt goes out, because the circuit is broken
  2. BIt stays on at the same brightness
  3. CIt gets brighter
  4. DIt flickers on and off

Answer: It goes out, because the circuit is broken

A series circuit has only one path. A gap anywhere stops the current everywhere.

17.Give two reasons why the lights and wall sockets in a house are wired in parallel, not in series.

Answer: Each appliance gets the full supply voltage, so it works properly; and each can be switched on and off on its own, so if one fails or is switched off the others keep working.

18.Adding more identical lamps in series to a battery makes all the lamps dimmer. Explain why, using resistance and current.

Answer: Each extra lamp adds resistance, so the total resistance increases and the current through the circuit decreases (I = V ÷ R). The supply voltage is also shared between more lamps, so each lamp gets less energy and glows less brightly.

Part E: Explain it

19.Compare series and parallel circuits. For each, describe how current and voltage behave, what happens if one lamp breaks, and give one place each type of circuit is used.[4 marks]

Model answer: Series: one path; the current is the same through every component; the supply voltage is shared between the components; if one lamp breaks, all the lamps go out; adding lamps makes each dimmer. Used where one switch must control everything, such as a switch or fuse in series with an appliance, or simple strings of decorative lights. Parallel: several branches; each branch gets the full supply voltage; the branch currents add to give the total current; if one lamp breaks, the others stay on at the same brightness. Used in household wiring and car headlights.

Success criteria:

  • Correct current behaviour for both circuits
  • Correct voltage behaviour for both circuits
  • Correct effect of a broken lamp in each
  • A sensible use of each
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