CIE iGCSE Co-Ordinated Science P4.2.2 Electric current Exam Style Questions Paper 4
Question
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P4.5.6 — The transformer / Transformer construction (Part (a)(i))
• Topic P4.5.6 — The transformer / Power loss calculations (Part (a)(ii))
• Topic P4.2.2 — Electric current / Direct vs alternating current (Part (b))
• Topic P4.5.2 — The a.c. generator / Slip rings (Part (c)(i))
• Topic P4.5.2 — The a.c. generator / e.m.f. graphs (Part (c)(ii) & (c)(iii))
▶️ Answer/Explanation
(a)(i) A step-up transformer consists of a soft iron core with a primary coil having fewer turns than the secondary coil.
The soft iron core provides a path for the magnetic field, and the turns ratio determines the voltage transformation. In a step-up transformer, the secondary coil has more turns, which increases the voltage.
(a)(ii) \( P = I^2R = 500^2 \times 0.60 = 250,000 \times 0.60 = 150,000 \text{ W} \)
Power loss in a transmission wire is calculated using \( P = I^2R \). The higher the current, the greater the energy lost as heat in the wires. Stepping up the voltage reduces the current and thus reduces power loss.
(b) Direct current (d.c.) flows in one direction only, while alternating current (a.c.) reverses direction periodically.
In d.c., the electric charge flows consistently from positive to negative. In a.c., the direction of charge flow changes repeatedly, typically in a sinusoidal pattern.
(c)(i) Slip rings maintain a continuous electrical connection to each side of the rotating coil, preventing the wires from twisting or tangling.
As the coil rotates, the slip rings rotate with it while the carbon brushes remain stationary, ensuring a constant electrical connection to the external circuit without wires becoming twisted.
(c)(ii) The graph should be a sine wave with positive and negative e.m.f. values, showing constant amplitude and a constant period.
As the coil rotates at constant speed, the e.m.f. varies sinusoidally: maximum positive, zero, maximum negative, and back to zero in one complete rotation. The amplitude and period remain constant.
(c)(iii) The amplitude of the e.m.f. doubles, and the period halves (the frequency doubles).
Doubling the rotation speed means the coil cuts through the magnetic field lines twice as fast, generating double the e.m.f. It also completes each rotation in half the time, so the period is halved.
Question
Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P4.2.2 — Electric current (Part (a)(i))
• Topic P4.2.3 — Voltage (Part (a)(ii))
• Topic P4.3.2 — Series and parallel circuits (Part (a)(iii), (a)(iv), (b)(i), (b)(ii))
• Topic P4.2.4 — Resistance (Part (a)(iv))
▶️ Answer/Explanation
(a)(i) Charge = 75 C.
\( Q = It = 0.25 \times 5.0 \times 60 = 75 \text{ C} \)
The equation linking charge (Q), current (I), and time (t) is:
\( Q = I \times t \)
Where Q is measured in coulombs (C), I in amperes (A), and t in seconds (s). The time given is 5.0 minutes, which must be converted to seconds: 5.0 × 60 = 300 seconds.
Therefore, \( Q = 0.25 \times 300 = 75 \text{ C} \).
(a)(ii) p.d. across the 20 Ω resistor = 3.0 V.
In a parallel circuit, the potential difference (voltage) across each branch is the same as the voltage across the power source. The cell provides 3.0 V across its terminals, so the p.d. across each resistor is also 3.0 V.
(a)(iii) Current in resistor R = 0.10 A.
\( I_R = 0.25 – 0.15 = 0.10 \text{ A} \)
In a parallel circuit, the total current supplied by the cell is the sum of the currents in each parallel branch:
\( I_{\text{total}} = I_{20\Omega} + I_R \)
\( 0.25 = 0.15 + I_R \)
\( I_R = 0.25 – 0.15 = 0.10 \text{ A} \)
(a)(iv) Total resistance = 12 Ω.
\( R_{\text{total}} = \frac{V}{I_{\text{total}}} = \frac{3.0}{0.25} = 12 \text{ Ω} \)
The total resistance of the circuit can be calculated using Ohm’s law:
\( R_{\text{total}} = \frac{V}{I_{\text{total}}} = \frac{3.0}{0.25} = 12 \text{ Ω} \)
(b)(i) The current in the cell decreases.
Explanation: Removing the branch containing resistor R increases the total resistance of the circuit. Since the voltage of the cell remains constant, the total current decreases (from \( I = V/R \)).
When the branch containing resistor R is removed, the circuit now consists only of the 20 Ω resistor. The total resistance of the circuit increases (from the combined parallel resistance to just the 20 Ω resistor). Since the voltage is constant, the total current from the cell decreases (Ohm’s law: \( I = V/R \)).
(b)(ii) The current in the 20 Ω resistor remains unchanged.
Explanation: The potential difference (p.d.) across the 20 Ω resistor remains the same (3.0 V, as it is still directly connected across the cell terminals). Since the resistance of the 20 Ω resistor is unchanged, the current through it remains \( I = V/R = 3.0/20 = 0.15 \text{ A} \).
The 20 Ω resistor is still directly connected across the cell terminals, so the potential difference across it remains 3.0 V. Since its resistance is unchanged, the current through it remains \( I = 3.0/20 = 0.15 \text{ A} \). The change in total current only affects the branch that was removed.