CBSE · Class 10 · Science

Class 10 Electricity Mock Test (CBSE)

6 practice questions from Electricity, in the style of your board paper, each with a worked answer and the marks for every step. Then take a full chapter test on paper, snap the pages and get every step marked.

Topics in Electricity

  • Electric current and circuit
  • Electric potential and potential difference
  • Circuit diagram
  • Ohm's law
  • Factors on which the resistance of a conductor depends
  • Resistors in series
  • Resistors in parallel
  • Heating effect of electric current and its applications
  • Electric power

Electricity practice questions with answers

  1. Q1Multiple choice1 mark
    Three resistors of 2 Ω2\ \Omega, 3 Ω3\ \Omega and 5 Ω5\ \Omega are connected in series with a cell and a plug key. The equivalent resistance of the combination is:
    1. (a)103 Ω\frac{10}{3}\ \Omega
    2. (b)6 Ω6\ \Omega
    3. (c)8 Ω8\ \Omega
    4. (d)10 Ω10\ \Omega
    Show answer and step marks
    1. Correct option: (d) 10 Ω10\ \Omega+1
    Answer: (d)
  2. Q2Multiple choice1 mark
    Two resistors of resistances 6 Ω6\ \Omega and 3 Ω3\ \Omega are connected in parallel. The equivalent resistance of the combination is:
    1. (a)9 Ω9\ \Omega
    2. (b)2 Ω2\ \Omega
    3. (c)0.5 Ω0.5\ \Omega
    4. (d)18 Ω18\ \Omega
    Show answer and step marks
    1. Correct option: (b) 2 Ω2\ \Omega+1
    Answer: (b)
  3. Q3Assertion and reason1 mark
    Assertion (A): In a series circuit, the same current flows through every resistor. Reason (R): Current is the rate of flow of charge, which is the same at all points along a single path.
    1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
    2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
    3. (c)Assertion (A) is true but Reason (R) is false.
    4. (d)Assertion (A) is false but Reason (R) is true.
    Show answer and step marks
    1. Correct option: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).+1
    Answer: (a)
  4. Q4Assertion and reason1 mark
    Assertion (A): Tungsten metal is used for making filaments of incandescent electric lamps. Reason (R): Tungsten has a very high melting point and a high resistivity.
    1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
    2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
    3. (c)Assertion (A) is true but Reason (R) is false.
    4. (d)Assertion (A) is false but Reason (R) is true.
    Show answer and step marks
    1. Correct option: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).+1
    Answer: (a)
  5. Q5Short answer2 marks
    (a) Define the potential difference between two points in an electric circuit. [1] (b) Write the SI unit of potential difference in terms of the joule and the coulomb. [1]
    Show answer and step marks
    1. (a) Potential difference between two points is the work done to move a unit positive charge from one point to the other: V=WQV = \dfrac{W}{Q}.+1
    2. (b) The SI unit is the volt (V), where 1 V=1 J C−11\ \text{V} = 1\ \text{J C}^{-1}.+1
    Answer: (a) Potential difference is the work done per unit positive charge moved between the two points, V=W/QV = W/Q. (b) Volt (V), with 1 V=1 J C−11\ \text{V} = 1\ \text{J C}^{-1}.
  6. Q6Short answer3 marks
    A timing buzzer at the side of a sports ground is connected across a 12 V12\ \text{V} battery. When the buzzer is switched on, a steady current of 2.5 A2.5\ \text{A} flows through it for 60 s60\ \text{s}. Calculate the work done by the battery in moving the charge through the buzzer.
    Show answer and step marks
    1. Charge passed: Q=I×t=2.5 A×60 s=150 CQ = I \times t = 2.5\ \text{A} \times 60\ \text{s} = 150\ \text{C}+1
    2. Work done is potential difference times charge: W=V×QW = V \times Q+1
    3. W=12 V×150 C=1800 JW = 12\ \text{V} \times 150\ \text{C} = 1800\ \text{J}+1
    Answer: 1800 J1800\ \text{J}

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