The problems are on the current and voltage worksheet. Try them on your own first, then compare with the answers below.
Quick answer: An ammeter measures electric current in amperes and is connected in series, in the loop. These worked answers correct the 20 problems of the grade 7 science worksheet on current and voltage, from quantities and units to an overloaded extension lead: each solution shows the formula, the substitution and the final answer with its unit.
Full answers are given for every problem, with the rule used at each step. When a calculation uses a rule, check that you picked the right one for a series or a parallel circuit. The key result of each step is in bold.
Level: Grade 7 science (Year 8 · Class 7) · Curriculum: MS-PS2-3
Problem 1 Quantities and units
1. The ampere, A.2. The volt, V.
3. Current is measured with an ammeter and voltage with a voltmeter.
Problem 2 Meter connections
1. In series: in the loop, so that all the current passes through it.2. In parallel: across the component.
3. The ammeter.
Problem 3 Changing units
1. 350 ÷ 1000 = 0.350 A.2. 0.045 × 1000 = 45 mA.
3. 2.4 × 1000 = 2400 mA.
Problem 4 Current in a series loop
1. 0.40 A.2. 0.40 A.
3. No. The current is the same everywhere because the charge has only one path and none is used up. The bulbs transfer energy, not charge.
Problem 5 Voltage shared in series
1. 6.0 V − 2.0 V = 4.0 V.2. They add up to the supply voltage, 6.0 V.
Problem 6 Reading an ammeter scale
1. The pointer is at the 13th mark of 0.05 A: 13 × 0.05 = 0.65 A.2. 0.65 × 1000 = 650 mA.
Problem 7 Voltage in parallel
1. 4.5 V across each bulb.2. Each bulb is connected straight across the supply, so each has the supply voltage.
3. Nothing changes: the other two bulbs still have 4.5 V.
Problem 8 Branch currents
1. 0.35 A + 0.20 A = 0.55 A.2. In a parallel circuit the branch currents add up to the main current.
Problem 9 The missing branch
1. 1.20 A − 0.45 A = 0.75 A.2. 0.75 × 1000 = 750 mA.
Problem 10 Three components in series
1. 9.0 V − 3.2 V − 2.8 V = 3.0 V.2. 0.15 A, since the current is the same everywhere in a series circuit.
Problem 11 Faulty meters
1. The ammeter is connected in parallel across the bulb, and the voltmeter is connected in series in the loop.2. The ammeter gives a very easy path, so the bulb goes out and a large current may damage the meter or the cell. It acts as a short circuit.
3. Put the ammeter in the loop and the voltmeter across the bulb.
Problem 12 Charger currents
1. 1500 ÷ 1000 = 1.5 A.2. 1.5 A + 0.50 A = 2.0 A.
3. 2.0 × 1000 = 2000 mA.
Problem 13 Current and voltage for a lamp
1. 0.42 A.2. 4.0 V.
3. 0.50 A − 0.32 A = 0.18 A.
Problem 14 Motorbike lamps
1. 12 V across each lamp, because they are in parallel.2. 2.5 A + 0.4 A = 2.9 A.
3. In the main wire from the battery, in series with the battery.
Problem 15 Solar panel readings
1. (0.32 + 0.54 + 0.71 + 0.83) ÷ 4 = 2.40 ÷ 4 = 0.60 A.2. 0.83 × 1000 = 830 mA.
3. Increases: 0.22 A, 0.17 A, 0.12 A. The largest is 0.22 A (between 08:00 and 09:00).
Problem 16 A series and parallel circuit
1. 9.0 V − 4.0 V = 5.0 V.2. 5.0 V, the same as L2 because they are in parallel.
3. 0.60 A − 0.35 A = 0.25 A.
4. 0.60 A, because all of the current passes through L1.
Problem 17 A voltmeter in the wrong place
1. About 3.0 V, very nearly the cell voltage.2. A voltmeter has a very high resistance, so it lets almost no current through the loop.
3. About 0 V across the bulb, because with almost no current the bulb takes almost none of the voltage.
Problem 18 Do the readings agree?
1. 0.62 + 0.48 + 0.35 = 1.45 A.2. 1.50 − 1.45 = 0.05 A.
3. Greatest total uncertainty = 3 × 0.02 + 0.02 = 0.08 A. The difference, 0.05 A, is smaller than this, so the readings agree with the rule.
Problem 19 Sharing 12 volts
1. 12.0 V − 2.4 V = 9.6 V.2. Let V₂ = x. Then 2x + x = 9.6 V, so x = 3.2 V. V₂ = 3.2 V and V₁ = 6.4 V.
3. 6.4 + 3.2 + 2.4 = 12.0 V, which equals the supply.
Problem 20 An overloaded extension lead
1. 1.2 A + 0.8 A + 1.5 A = 3.5 A.2. 3.5 A is greater than the 3.0 A rating, so the fuse will blow.
3. 1.2 A + 1.5 A = 2.7 A, which is less than 3.0 A, so the fuse is safe.
