Electric Circuits: Current, Voltage & Resistance, Series & Parallel Circuits, Power & Energy, Household Electricity & Safety, Cost of Electricity (Lessons 5–11)
8 questions — select the best answer
Which of the following correctly describes the relationship between current, voltage and resistance for an ohmic conductor at constant temperature?
Three resistors ($2\ \Omega$, $3\ \Omega$, $5\ \Omega$) are connected in series to a 20 V battery. What is the current in the circuit?
Two resistors ($6\ \Omega$ and $12\ \Omega$) are connected in parallel. What is their total resistance?
In a parallel circuit, which quantity is the same across all branches?
A 2400 W heater operates from 240 V mains. What current does it draw?
Which safety device is designed to prevent electrocution by detecting current imbalance between active and neutral wires?
A 100 W incandescent globe and a 10 W LED globe both run for 100 hours. The incandescent globe uses:
In a combined circuit, a $4\ \Omega$ resistor is in series with a parallel combination of two $4\ \Omega$ resistors. The total resistance is:
9. A circuit contains a $3\ \Omega$ resistor in series with a parallel combination of a $6\ \Omega$ and a $12\ \Omega$ resistor. The circuit is connected to a 24 V battery. (a) Calculate the total resistance. (b) Calculate the total current. (c) Calculate the current through the $6\ \Omega$ resistor. 4 MARKS
10. A 1500 W hair dryer operates from 240 V mains. (a) Calculate the current it draws. (b) Calculate its operating resistance. (c) The hair dryer is connected to an extension cord with total resistance $0.3\ \Omega$. Calculate the power dissipated as heat in the cord and explain why this is a potential safety hazard. 5 MARKS
11. A student designs a circuit for a model house with a 12 V battery, four 6 V light globes, and a switch. The student connects all four globes in series to the battery. Evaluate this design with reference to voltage division, brightness, and the effect of one globe failing. Suggest and justify an improved design. 6 MARKS
Q1: B — Ohm's Law: $I = V/R$, so $I \propto V$ at constant $R$.
Q2: C — $R_{\text{total}} = 2 + 3 + 5 = 10\ \Omega$; $I = 20/10 = 2.0\ \text{A}$.
Q3: A — $R_{\text{total}} = (6 \times 12)/(6 + 12) = 72/18 = 4\ \Omega$.
Q4: D — Voltage is constant across all branches in parallel.
Q5: B — $I = P/V = 2400/240 = 10\ \text{A}$.
Q6: C — RCDs detect current imbalance and protect against electrocution.
Q7: A — $E = Pt$; same time, so energy ratio = power ratio = 100/10 = 10.
Q8: D — Parallel pair: $R_p = (4 \times 4)/(4 + 4) = 2\ \Omega$. Total: $4 + 2 = 6\ \Omega$.
Q9 (4 marks): (a) $1/R_p = 1/6 + 1/12 = 3/12 = 1/4$, so $R_p = 4\ \Omega$. $R_{\text{total}} = 3 + 4 = 7\ \Omega$. (b) $I_{\text{total}} = 24/7 = 3.43\ \text{A}$. (c) $V_p = I_{\text{total}} \times R_p = 3.43 \times 4 = 13.7\ \text{V}$. $I_6 = 13.7/6 = 2.29\ \text{A}$.
Q10 (5 marks): (a) $I = P/V = 1500/240 = 6.25\ \text{A}$. (b) $R = V/I = 240/6.25 = 38.4\ \Omega$ (or $R = V^2/P = 240^2/1500 = 38.4\ \Omega$). (c) $P_{\text{cord}} = I^2R = (6.25)^2 \times 0.3 = 39.06 \times 0.3 = 11.7\ \text{W}$. This is hazardous because the extension cord dissipates nearly 12 W as heat. If the cord is thin, coiled, or poorly ventilated, this heat can raise the insulation temperature above safe limits, potentially melting the insulation and causing a fire. The cord should be rated for at least 10 A and should be fully unwound during use.
Q11 (6 marks): The series design is fundamentally flawed for several reasons. First, in a series circuit, voltage divides among components according to their resistance. With four identical 6 V globes in series connected to 12 V, each globe receives only $12/4 = 3$ V — half their rated voltage. Globes operate at reduced brightness (well below optimal) when under-volted, and may not glow at all if the voltage is too low. Second, a series circuit has a single current path, so if one globe fails (burns out, creating an open circuit), all globes go out. This is unacceptable for a house lighting system. Third, adding or removing globes changes the voltage across each remaining globe, causing unpredictable brightness changes.
An improved design connects the globes in parallel across the 12 V battery. In parallel, each globe receives the full 12 V — but this exceeds the 6 V rating and would burn out the globes immediately. The correct solution is to connect pairs of globes in series, then connect these pairs in parallel across the battery. Each series pair receives $12/2 = 6$ V (rated voltage), and the parallel arrangement ensures that if one pair fails, the other pair continues to operate. Alternatively, use four 12 V globes in parallel. This design ensures correct operating voltage, independent operation, and reliability.
Tick when you have finished all questions and checked your answers.