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Electricity

CBSE Class 10 · Science · Chapter 11

Past board exam questions

CBSE 2027 sample paper · 1 mark

An electrical appliance rated 1000 W is operated at 250 V. What current does it draw?

  1. 0.25 A
  2. 2.5 A
  3. 4.0 A
  4. 10 A
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4.0 A

CBSE 2027 sample paper · 2 marks

I. State Joule’s Law of heating. II. What happens to the heat produced in a conductor if the resistance is doubled while keeping other variables constant?

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I. Joule's law of heating: the heat produced in a resistor is (i) directly proportional to the square of the current for a given resistance, (ii) directly proportional to the resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor: H = I²Rt, where I is the current and R the resistance. II. For the same current I and time t: H' = I²(2R)t = 2I²Rt = 2H, i.e. the heat produced doubles.

CBSE 2027 sample paper · 3 marks

(Circuit: points A and B are joined directly by a 2 Ω resistor. A second path between A and B runs through two 1 Ω resistors in series, forming the two upper sides of a triangle above the 2 Ω resistor. From B the wire continues to the right, where a 3 Ω resistor is connected in parallel with another 3 Ω resistor (drawn in a lower V-shaped branch). From the far end of this pair the wire goes down and along the bottom through a battery V = 15 V, then up through a 5 Ω resistor back to A.) The circuit described above has a 15 V battery connected to several resistors. Using the given circuit: A. Find the total current supplied by the battery. B. Calculate the potential difference between points A and B.

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A. The two 1 Ω resistors are in series: 1 + 1 = 2 Ω; this combination is in parallel with the 2 Ω resistor between A and B: (1)/(RAB)=(1)/(2)+(1)/(2)=1, so RAB=1 Ω. The two 3 Ω resistors are in parallel: (1)/(R)=(1)/(3)+(1)/(3)=(2)/(3), R = 1.5 Ω. 5 Ω, 1 Ω and 1.5 Ω are in series: R_s = 5 + 1 + 1.5 = 7.5 Ω. V = IR: 15 V = I × 7.5 Ω, so I = 2 A. B. The whole current of 2 A passes between A and B: VAB=I× RAB=2×1=2 V.

CBSE 2027 sample paper · 2 marks

I. Define resistivity. II. Justify with an example why resistivity is an intrinsic property of a material.

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I. The resistivity (ρ) of a material is the resistance of a wire of that material having unit length and unit cross-sectional area. From R=ρ(L)/(A), ρ=(RA)/(L), where R is the resistance of a wire of length L and cross-sectional area A. SI unit: Ω m. II. Resistivity depends only on the nature of the material (and temperature), not on its dimensions. Example: two copper wires of different lengths and thicknesses have different resistances, but when ρ is calculated as R × A / L for each, the same value (about 1.62 × 10⁻⁸ Ω m) is obtained.

CBSE 2026 board paper · 1 mark

The correct way to connect an ammeter and a voltmeter in an electric circuit is :

  1. Ammeter in parallel and voltmeter in series
  2. Ammeter and voltmeter both in parallel
  3. Ammeter in series and voltmeter in parallel
  4. Ammeter and voltmeter both in series
Show answer
Ammeter in series and voltmeter in parallel

CBSE 2026 board paper · 1 mark

In an electric circuit carrying current I, a 2 Ω resistor is replaced by an 8 Ω resistor without changing the connected (ideal) battery. Then :

  1. current becomes (I)/(4) and potential difference (V)/(4)
  2. current becomes (I)/(4) and potential difference remains V
  3. current becomes 4I and potential difference remains V
  4. current becomes 4I and potential difference (V)/(4)
Show answer
current becomes (I)/(4) and potential difference remains V

CBSE 2026 board paper · 2 marks

(A) A fuse in electric circuit is rated 4 A. Can it be used with an electric heater of rating 2 kW, 200 V ? Explain your answer.

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Current drawn by the heater: I = (P)/(V) = (2000)/(200) = 10 A. This is much more than the 4 A rating of the fuse, so the fuse wire would melt and break the circuit. Hence the fuse cannot be used with this heater.

CBSE 2026 board paper · 2 marks

Calculate the total resistance of the given circuit and find the total current flowing in it. Circuit: one terminal of a 12 V battery is joined through R₁ = 1 Ω to a junction P; the other terminal is joined to a junction Q. From P to Q there are two paths: one through R₂ = 10 Ω, and the other through R₃ = 6 Ω followed by R₄ = 4 Ω.

Show answer
R₃ and R₄ are in series: R₃ + R₄ = 6 Ω + 4 Ω = 10 Ω. This 10 Ω is in parallel with R₂ = 10 Ω: (1)/(Rp) = (1)/(10) + (1)/(10) = (2)/(10), so R_p = 5 Ω. R₁ is in series with this combination: total resistance R = 1 Ω + 5 Ω = 6 Ω. Total current I = V/R = 12 V / 6 Ω = 2 A.