CBSE Class 10 Science 2026: question 39(A)

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Step-by-step solution

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1. (i)(I) $R_2$ is in parallel with the 4.0 Ω resistor, so both have the same potential difference.
   $V=I_{1}R=1.5\times 4=6\text{V}$

   1 mark

2. (II) Current through $R_2$: $I_2=I-I_1=2.0-1.5=0.5\text{A}$.
   $R_2=\frac{V}{I_2}=\frac{6}{0.5}=12\Omega$

   1 mark

3. (III) Potential difference across the 2.0 Ω resistor: $V=IR=2\times 2=4\text{V}$.
   Potential difference across $R_1$: $12-(6+4)=2\text{V}$.
   $R_1=\frac{V}{I}=\frac{2}{2}=1\Omega$

   1 mark

4. (ii) Resistance of a conductor depends on:

   • its length ($R\propto l$);
   • its area of cross-section ($R\propto \frac{1}{A}$);
   • the nature of its material (and its temperature).
   1 mark

5. Combining the two proportionalities: $R\propto \frac{l}{A}$, so $R=\rho \frac{l}{A}$.
   Here $\rho$ is the proportionality constant called the resistivity of the material.

   1 mark

Answer: (I) 6 V; (II) R₂ = 12 Ω; (III) R₁ = 1 Ω; (ii) R depends on length, area of cross-section and material, giving R = ρl/A.

Revise the chapter

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