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Science · 2023 · 4 marks
CBSE 2023 · Region 2 · Set 1 · Q39
Consider the following electrical circuit diagram in which nine identical resistors of 3 Ω each are connected as shown. If the reading of the ammeter…
Consider the following electrical circuit diagram in which nine identical resistors of $\displaystyle 3 \Omega$ each are connected as shown. If the reading of the ammeter $\displaystyle \mathrm{A}_{1}$ is $\displaystyle 1$ ampere, answer the following questions :
(a)
What is the relationship between the readings of $\displaystyle \mathrm{A}_{1}$ and $\displaystyle \mathrm{A}_{3}$ ? Give reasons for your answer.
(b)
What is the relationship between the readings of $\displaystyle \mathrm{A}_{2}$ and $\displaystyle \mathrm{A}_{3}$ ?
(c)
Determine the reading of the voltmeter $\displaystyle \mathrm{V}_{1}$.
OR
(c)
Find the total resistance of the circuit.
Marking-scheme solution
(a) • Both have same reading/ \(\displaystyle \mathrm{A}_{1}=\mathrm{A}_{3}\)
Both are connected in series
(b) Reading of \(\displaystyle \mathrm{A}_{2}=\frac{1}{4} \mathrm{~A}\) as current is equally divided in the four identical resitors . /Reading of \(\displaystyle \mathrm{A}_{2}=\frac{1}{4}\) times Reading of \(\displaystyle \mathrm{A}_{3} . / \mathrm{A}_{2}=0.25 \mathrm{~A} / \mathrm{A}_{2}<\mathrm{A}_{3}\)
(c)
\(\displaystyle \frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \quad / \quad \mathrm{R}_{\mathrm{p}}=\frac{R}{n}\)
\(\displaystyle \frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{3 \Omega}+\frac{1}{3 \Omega}\)
V = I R
\(\displaystyle \mathrm{V}_{1}=1 \mathrm{~A} \times \frac{3}{2} \Omega=\frac{3}{2} \mathrm{~V}=1 \cdot 5 \mathrm{~V}\)
\(\displaystyle \frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{3 \Omega}+\frac{1}{3 \Omega}\)
\(\displaystyle \frac{1}{\mathrm{R}_{\mathrm{p}}}=\frac{1}{3 \Omega}+\frac{1}{3 \Omega}+\frac{1}{3 \Omega}\)
\(\displaystyle \therefore \mathrm{R}_{\mathrm{p}_{2}}=1 \Omega\)
\(\displaystyle \therefore \mathrm{R}=\mathrm{R}_{\mathrm{p}_{1}}+\mathrm{R}_{\mathrm{p}_{2}}+\mathrm{R}_{\mathrm{p}_{3}}=\left(\frac{3}{2}+1+\frac{3}{4}\right) \Omega=\frac{13}{4} \Omega / 3.25 \Omega\)
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CBSE Class 10 Science past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.