CIE iGCSE Co-Ordinated Science P4.2.2 Electric current Exam Style Questions Paper 3
Question
The energy is stored in a battery in the car.
The battery supplies a current of 96 A at 120 V to the motor that drives the car.
The tyre exerts a pressure of 20 N/cm² on the road.
Calculate the force exerted by the tyre on the road.
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P6.1 — The Solar System (Part (a))
• Topic P4.2.5 — Electrical energy and electrical power (Part (b)(i))
• Topic P4.2.2 — Electric current (Part (b)(ii))
• Topic P1.5.2 — Turning effect of forces (Part (c)(i))
• Topic P1.7 — Pressure (Part (c)(ii))
▶️ Answer/Explanation
(a)(i) Gravitational attraction
The Sun’s gravitational field provides the centripetal force that keeps the Earth in its nearly circular orbit around the Sun.
(a)(ii) Milky Way
The Sun is one of approximately 100–400 billion stars in the Milky Way galaxy, which is a spiral galaxy about 100,000 light-years in diameter.
(b)(i) Energy = \(I \times V \times t = 96 \times 120 \times 900 = 10,368,000 \text{ J} \approx 10,000,000 \text{ J}\)
The electrical energy transferred is calculated using the formula \(E = IVt\), where current is in amperes, voltage in volts, and time in seconds. The unit of energy is the joule (J).
(b)(ii) Direct current (d.c.) flows in one direction only, whereas alternating current (a.c.) periodically changes direction.
In a direct current circuit, electrons flow continuously in the same direction. In an alternating current circuit, the direction of electron flow reverses periodically (e.g., 50 times per second in mains electricity).
(c)(i) moment = \(300 \times 0.40 = 120 \text{ N m}\)
The moment of a force is calculated using the equation \(\text{moment} = \text{force} \times \text{perpendicular distance from the pivot}\). The unit is newton-metre (N m).
(c)(ii) force = \(20 \times 180 = 3600 \text{ N}\)
Pressure is defined as force per unit area. The force can be calculated by multiplying pressure by area: \(\text{force} = \text{pressure} \times \text{area}\).
Question
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P4.2.2 — Electric current (Part (a))
• Topic P1.6.3 — Energy transfers (Part (b))
• Topic P3.2.2 — Refraction of light (Part (c))
• Topic P3.1 — General Properties of Waves (Part (d))
▶️ Answer/Explanation
(a) current; volts; current
A flow of electrons is a current.
e.m.f. is measured in volts (V).
Resistance \(R = \dfrac{V}{I}\), so it is voltage divided by current.
(b) 360 J
Energy is conserved, so input energy = useful output + wasted energy.
Wasted energy = \(400 \, \text{J} – 40 \, \text{J} = 360 \, \text{J}\).
Most of the electrical energy supplied to a filament lamp is wasted as heat.
(c)(i) Ray bends towards the normal on entering the glass
Light slows down as it enters the denser glass block.
This causes the ray to bend towards the normal line at the point of entry.
The refracted ray then continues in a straight line through the glass.
(c)(ii) i and r labelled correctly
The angle of incidence, \(i\), is measured between the incident ray and the normal.
The angle of refraction, \(r\), is measured between the refracted ray and the normal.
Since the ray bends towards the normal, \(r\) is smaller than \(i\).
(d)(i) One wavelength shown correctly
A wavelength, \(\lambda\), is the distance from one point on a wave to the identical point on the next wave.
It is typically measured from one peak to the next peak (or one trough to the next trough).
The double-headed arrow labelled W should span exactly one full wave cycle.
(d)(ii) Amplitude shown correctly
Amplitude is the maximum displacement of a point on the wave from its undisturbed (rest) position.
It is measured vertically from the midline to the peak (or trough), not peak-to-trough.
The double-headed arrow labelled A should be drawn from the centre line up to the maximum point.