ScienceCBSE Class 10 · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 10 Science sample paper 2024-25: question 36(A)

From the official CBSE sample question paper for 2024-25, section D, question 36(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q36(A) · 5 marks · Long answerElectricity

The above circuit is a part of an electrical device. Use the information given in the question to calculate the following.
(i) Potential Difference across $R_{2}$ .
(ii) Value of the resistance $R_{2}$ .
(iii) Value of resistance $R_{1}$

(For visually impaired students: A. (i) State the law and write the formula connecting the electric current flowing through a conductor and voltage applied across it. (ii) In a household five fans each of 100W are used for 4 hours and an electric press of 500W for 2 hours every day. Calculate the cost of using the fans and electric press for 60 days if the cost of 1 unit of electrical energy is Rs. 6.5.)

Figure in the paper: 12 V battery in series with resistor R1 and a 2.0 Ω resistor, and a parallel combination of a 4.0 Ω resistor (current 1.5 A) and R2. Total current 2.0 A. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

Resistors in parallel always share the same potential difference.

Hint 2

The total current splits between the parallel branches. Find the share that goes through R₂.

Hint 3

In a series circuit, the battery voltage equals the sum of the p.d.s across each part.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) The 4.0 Ω resistor and $R_{2}$ are in parallel, so the potential difference across them is the same.
½ mark
Using $V = I R$ for the 4.0 Ω resistor:
$V = 1.5 A \times 4.0 Ω = 6 V$
So the p.d. across $R_{2}$ is 6 V.
½ mark
(ii) The total current of 2.0 A divides between the two parallel branches. The 4.0 Ω resistor takes 1.5 A.
½ mark
Current through $R_{2}$ :
$I_{2} = 2.0 - 1.5 = 0.5 A$
Apply Ohm's law to $R_{2}$ : $6 V = 0.5 A \times R_{2}$ .
½ mark
$R_{2} = \frac{6}{0.5} = 12 Ω$
½ mark
(iii) $R_{1}$ , the 2.0 Ω resistor and the parallel combination are in series with the 12 V battery. So:
$V_{R_{1}} = 12 V - V_{parallel} - V_{2 Ω}$
½ mark
P.d. across the 2.0 Ω resistor $= 2.0 A \times 2.0 Ω = 4 V$
P.d. across the parallel part $= 6 V$ (from part (i))
½ mark
$V_{R_{1}} = 12 - 6 - 4 = 2 V$
½ mark
$R_{1}$ is in the main line, so the full 2.0 A flows through it.
½ mark
$R_{1} = \frac{V}{I} = \frac{2 V}{2 A} = 1 Ω$
*(Check: the total resistance should be $\frac{12}{2} = 6 Ω$ . Here $R_{1} + \frac{4 \times 12}{4 + 12} + 2 = 1 + 3 + 2 = 6 Ω$ ✓)*
½ mark

Answer: (i) 6 V; (ii) R₂ = 12 Ω; (iii) R₁ = 1 Ω

Where the marks are

States that the p.d. is the same across parallel resistors (0.5 mark)
Calculates 1.5 × 4 = 6 V (0.5 mark)
Finds the current in R₂ as 0.5 A using the total current (0.5 mark)
Applies Ohm's law to R₂ (0.5 mark)
Obtains R₂ = 12 Ω (0.5 mark)
Writes the p.d. relation for the series circuit (0.5 mark)
Finds the p.d. across 2 Ω (4 V) and uses 6 V across the parallel part (0.5 mark)
Obtains 2 V across R₁ (0.5 mark)
Notes that the current through R₁ is 2 A (0.5 mark)
Obtains R₁ = 1 Ω (alternatively, via total resistance 6 Ω) (0.5 mark)

Revise the chapter

Chapter: Electricity (notes, formulas, where marks are lost and every past question)
Topic: Ohm's Law

Practise this chapterBoard-style questions from Electricity, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Science sample paper 2024-25: question 36(A)

Hints

Step-by-step solution

Revise the chapter

More questions from Electricity