This end-of-module quiz assesses your understanding of all three inquiry questions: electrostatics (Lessons 1–4), electric circuits (Lessons 5–11), and magnetism (Lessons 12–16). It is designed to simulate exam conditions and identify areas for revision before your Module 4 assessment.
1. Two identical metal spheres carry charges of $+4\ \mu\text{C}$ and $-8\ \mu\text{C}$. After they touch and are separated, the charge on each sphere is: 1 MARK
2. Three resistors ($2\ \Omega$, $4\ \Omega$, $6\ \Omega$) are connected in series to a 12 V battery. The current in the circuit is: 1 MARK
3. A wire carries current vertically upward. What is the direction of the magnetic field at a point directly to the west of the wire? 1 MARK
4. A $60\ \text{W}$ light globe operating at $240\ \text{V}$ draws a current of approximately: 1 MARK
5. In a household circuit, a $2400\ \text{W}$ heater and a $120\ \text{W}$ television are connected in parallel to $240\ \text{V}$. The total current drawn from the supply is: 1 MARK
6. The magnetic field inside a long solenoid is approximately: 1 MARK
7. Two resistors ($6\ \Omega$ and $12\ \Omega$) are connected in parallel across a $12\ \text{V}$ battery. The total power dissipated is: 1 MARK
8. A circuit contains a $12\ \text{V}$ battery, a $4\ \Omega$ resistor, and a $12\ \Omega$ resistor connected in series with a $6\ \Omega$ resistor connected in parallel across the $12\ \Omega$ resistor. 5 MARKS
9. A student connects three resistors ($2\ \Omega$, $4\ \Omega$, and $6\ \Omega$) to a $12\ \text{V}$ battery. She first connects all three in series and measures the current. She then reconfigures the circuit so that the $4\ \Omega$ and $6\ \Omega$ resistors are in parallel, with this combination in series with the $2\ \Omega$ resistor. (a) Calculate the current supplied by the battery in each configuration. (b) Calculate the total power dissipated in each configuration. (c) Explain why the parallel-series configuration dissipates more power, with reference to the equivalent resistance and the conservation of energy. 5 MARKS
(a) $1/R_p = 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4$. Therefore $R_p = 4\ \Omega$.
(b) $R_{\text{total}} = 4 + 4 = 8\ \Omega$.
(c) $I = V/R = 12/8 = 1.5$ A.
(d) $V_4 = IR = (1.5)(4) = 6.0$ V.
(e) Voltage across parallel combination = $12 - 6.0 = 6.0$ V. $P_6 = V^2/R = (6.0)^2/6 = 6.0$ W. Alternatively: current through $6\ \Omega$ = $V/R = 6.0/6 = 1.0$ A. $P = I^2R = (1.0)^2(6) = 6.0$ W.
(a) Series configuration: $R_{\text{total}} = 2 + 4 + 6 = 12\ \Omega$. $I = V/R = 12/12 = 1.0$ A.
Parallel-series configuration: $1/R_p = 1/4 + 1/6 = 3/12 + 2/12 = 5/12$, so $R_p = 2.4\ \Omega$. $R_{\text{total}} = 2 + 2.4 = 4.4\ \Omega$. $I = V/R = 12/4.4 = 2.73$ A (or $30/11$ A).
(b) Series power: $P = VI = 12 \times 1.0 = 12$ W (or $P = I^2R = (1.0)^2(12) = 12$ W).
Parallel-series power: $P = VI = 12 \times 2.73 = 32.7$ W (or $P = V^2/R = 144/4.4 = 32.7$ W).
(c) The parallel-series configuration dissipates more power because its equivalent resistance ($4.4\ \Omega$) is lower than the series resistance ($12\ \Omega$). From $P = V^2/R$, lower resistance at the same voltage means higher power. By the law of conservation of energy, the power dissipated in the resistors equals the power supplied by the battery. The battery supplies more current in the parallel-series configuration because the total resistance is lower, so more energy is transferred per second from chemical energy in the battery to thermal energy in the resistors.