Chapter-14 Electricity — Online MCQ Test
PHYSICS · Grade 10 · IGCSE cambridge
Practice Chapter-14 Electricity with a free chapter-wise online MCQ test.
This chapter covers: Static electricity current voltage resistance series and parallel circuits and electrical safety..
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter-14 Electricity — Important Questions & Answers
What is the SI unit of electric current?
- A. Ampere (A)
- B. Volt (V)
- C. Ohm (Ω)
- D. Coulomb (C)
Answer: A. Ampere (A)
The ampere is the SI unit of electric current, defined as the flow of charge per unit time.
The ampere is the SI unit of electric current, defined as the flow of charge per unit time.
Static electricity is caused by the transfer of _____ between objects.
- A. protons
- B. electrons
- C. neutrons
- D. ions
Answer: B. electrons
Static electricity results from the accumulation of electrons (or loss of electrons) on an object's surface due to friction or contact.
Static electricity results from the accumulation of electrons (or loss of electrons) on an object's surface due to friction or contact.
In a series circuit, how does the current behave as it passes through multiple resistors?
- A. The current increases at each resistor
- B. The current remains the same throughout the circuit
- C. The current decreases at each resistor
- D. The current splits equally among resistors
Answer: B. The current remains the same throughout the circuit
In a series circuit, current is the same at all points because charge cannot accumulate or be lost; it flows sequentially through each component.
In a series circuit, current is the same at all points because charge cannot accumulate or be lost; it flows sequentially through each component.
A circuit contains two resistors: R₁ = 6Ω in series with a parallel combination of R₂ = 3Ω and R₃ = 6Ω. What is the total resistance of the circuit?
- A. 15Ω
- B. 8Ω
- C. 9Ω
- D. 12Ω
Answer: B. 8Ω
Parallel resistance of R₂ and R₃: 1/R_p = 1/3 + 1/6 = 3/6, so R_p = 2Ω. Total = R₁ + R_p = 6 + 2 = 8Ω.
Parallel resistance of R₂ and R₃: 1/R_p = 1/3 + 1/6 = 3/6, so R_p = 2Ω. Total = R₁ + R_p = 6 + 2 = 8Ω.
A heating element has a resistance of 20Ω. When connected to a 240V supply, it operates normally. However, if due to a fault the resistance suddenly increases to 30Ω at the same voltage, and the circuit has a 20A circuit breaker, which of the following correctly describes the outcome?
- A. The current increases, exceeding 20A, and the circuit breaker trips
- B. The current decreases to 8A, the breaker does not trip, but heating reduces
- C. The current increases to 30A, causing the breaker to trip
- D. The current remains constant due to the constant voltage supply
Answer: B. The current decreases to 8A, the breaker does not trip, but heating reduces
With R = 30Ω and V = 240V, current I = 240/30 = 8A, which is below the 20A limit. The breaker does not trip, but power output decreases from P = V²/R.
With R = 30Ω and V = 240V, current I = 240/30 = 8A, which is below the 20A limit. The breaker does not trip, but power output decreases from P = V²/R.