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NCERT Class 10 Science Electricity: In-text questions (page 185) Solutions

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of In-text questions (page 185) in NCERT's Class 10 Science book, chapter “Electricity”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

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How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.

Tap “why?” under a step to see the textbook rule it uses — from Shishya's fixed list of NCERT Class 9-10 rules, not written by AI for this page. A step without one has no rule from the list yet.

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  1. In-text (page 185) Q1

    Draw the circuit: a battery of three 2 V cells, resistors of 5 Ω, 8 Ω and 12 Ω and a plug key, all in series.

    1. Draw a rectangular loop to represent the circuit wire.
    2. Place a battery symbol on one side with three cells connected in series, labeling the total voltage as 6 V.
    3. Draw three resistor symbols (zig-zag lines) in series along the top wire, labeling them 5 Ω, 8 Ω, and 12 Ω.
      why?

      In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

    4. Draw a plug key symbol in the open position on the bottom wire to complete the series circuit.

    Answer: A series circuit diagram containing a 6 V battery, resistors of 5 Ω, 8 Ω, and 12 Ω, and a plug key.

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  2. In-text (page 185) Q2

    In that circuit, with an ammeter for the current and a voltmeter across the 12 Ω resistor, what do the meters read?

    1. The total resistance of the series circuit is 12 Ω + 3 Ω = 15 Ω.
      why?

      In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

    2. The current through the circuit is I = V ÷ R = 3.6 V ÷ 15 Ω = 0.24 A.
      why?

      Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

    3. The ammeter reads the total current, which is 0.24 A.
    4. The voltage across the 12 Ω resistor is V = I × R = 0.24 A × 12 Ω = 2.88 V.
      why?

      Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

    5. The voltmeter reads 2.88 V.

    Answer: Ammeter reads 0.24 A; Voltmeter reads 2.88 V

    Report an error in this solution

Other exercises in this chapter

← Electricity: notes, practice and all exercises

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