📚 CBSE Grade-10 Study Guide Open in the Study Guide single page app →
HomeScience

Science — CBSE Class 10 board question

Q1. [2]
Use Ohm's law to determine the potential difference across the 3 Ω resistor in the circuit shown in the following diagram when key is closed :
Previously asked in CBSE board exam
2024 31/2/1 Q25
Generated by claude-sonnet-4-6 · 2026-06-15 06:46 · grounding rag
Model Answer

The circuit has resistors 1 Ω, 2 Ω, and 3 Ω connected in series with a 2 V cell.

Step 1: Total resistance
$$R_s = 1 + 2 + 3 = 6 \text{ Ω}$$

Step 2: Current in circuit (Ohm's law)
$$I = \frac{V}{R_s} = \frac{2 \text{ V}}{6 \text{ Ω}} = \frac{1}{3} \text{ A}$$

Step 3: Potential difference across 3 Ω resistor
$$V_3 = I \times R = \frac{1}{3} \times 3 = \textbf{1 V}$$

Source: Chapter 11, Section 11.6.1 – Resistors in Series

---

Explanation
If a question refers to an image, map, graph or diagram that is not shown here, open the Study Guide single page app, go to Library and find the actual CBSE question paper. The original papers are also available on the CBSE website: cbse.gov.in.
Previous-year CBSE Grade 10 board exam questions, organised by subject and chapter, each with a model answer — free to read and print.