📚 CBSE Grade-10 Study Guide
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Science (086) — AI-generated practice question

AI-generated practice question — model-generated for extra practice, not a previous-year CBSE board question.

Q1. [5] medium thorough-understanding
[long_answer] Three resistors of 5 Ω, 10 Ω, and 15 Ω are connected in series across a 30 V supply. (a) Calculate the current through the circuit and the potential difference across each resistor. (b) Which resistor dissipates the most power? Justify using an appropriate formula. (c) If the 15 Ω resistor is now removed and the remaining two resistors stay connected in series to the same 30 V supply, how does the power dissipated in the 5 Ω resistor change? Support your answer with calculations.
Generated by claude-sonnet-4-6 · 2026-06-26 01:11 · grounding rag
Model Answer

(a) Current and potential difference:

Total resistance: $R_s = 5 + 10 + 15 = 30\ \Omega$

Current: $I = V/R_s = 30/30 = \mathbf{1\ A}$

Potential difference across each resistor (V = IR):

(b) Resistor dissipating most power:

Using $P = I^2R$ (current is same in series):

The 15 Ω resistor dissipates the most power. In a series circuit, since current is constant, $P \propto R$, so the largest resistor dissipates maximum power.

(c) Change in power dissipated in 5 Ω resistor:

New circuit: 5 Ω and 10 Ω in series, V = 30 V

New total resistance: $R_s = 5 + 10 = 15\ \Omega$

New current: $I' = 30/15 = 2\ \text{A}$

New power in 5 Ω: $P' = I'^2 \times 5 = 4 \times 5 = \mathbf{20\ W}$

Previously, $P = 1^2 \times 5 = 5\ \text{W}$. The power dissipated in the 5 Ω resistor increases from 5 W to 20 W (4 times) because removing the 15 Ω resistor reduces total resistance, increasing current.

Source: Chapter 11, Sections 11.6.1 and 11.8

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Explanation
Previous-year CBSE Grade 10 board exam questions, organised by subject and chapter, each with a model answer — free to read and print.