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Science · 2025 · 4 marks
CBSE 2025 · Region 3 · Set 1 · Q39
As shown in the diagram, an electric circuit consisting of an ammeter, a voltmeter, 4 cells of 1 · 5 V each, a plug key with a gap XY was set up.…
As shown in the diagram, an electric circuit consisting of an ammeter, a voltmeter, $\displaystyle 4$ cells of $\displaystyle 1 \cdot 5 \mathrm{~V}$ each, a plug key with a gap XY was set up. Voltmeter and ammeter readings were recorded in the observation table for four arrangements as given below : Arrangement No. $\displaystyle 1$ - only resistor $\displaystyle \mathrm{R}_{1}$ in gap XY Arrangement No. $\displaystyle 2$ - only resistor $\displaystyle \mathrm{R}_{2}$ in gap XY Arrangement No. $\displaystyle 3$ - Resistors $\displaystyle \mathrm{R}_{1}$ and $\displaystyle \mathrm{R}_{2}$ in series in gap XY Arrangement No. $\displaystyle 4$ - Resistors $\displaystyle \mathrm{R}_{1}$ and $\displaystyle \mathrm{R}_{2}$ in parallel in gap XY
Based on the observations, four V - I graphs A, B, C and D as shown in figure were drawn. Study these graphs.
(a)
Which one of the graphs represents the series combination of $\displaystyle \mathrm{R}_{1}$ and $\displaystyle \mathrm{R}_{2}$ ?
(b)
Which one of these graphs represents the parallel combination of $\displaystyle \mathrm{R}_{1}$ and $\displaystyle \mathrm{R}_{2}$ ?
(c)(i)
Show an arrangement of three resistors, each of resistance $\displaystyle 10 \Omega$, so that the combination has a resistance of $\displaystyle 15 \Omega$. Give justification for your answer.
OR
(c)
(ii)
A battery of $\displaystyle 6$ V is connected with a series combination of five resistors of $\displaystyle 0 \cdot 1 \Omega, 0 \cdot 2 \Omega, 0 \cdot 3 \Omega, 0 \cdot 4 \Omega$ and $\displaystyle 0 \cdot 5 \Omega$. How much current would flow through the $\displaystyle 0 \cdot 3 \Omega$ resistor ? Justify your answer.
Marking-scheme solution
(a) Graph A
(b) Graph D,
(c) (i)
\[\mathrm{R}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}+R_{3}
\]
\[\mathrm{R}=\left(\frac{10 \times 10}{10+10}+10\right) \Omega=5 \Omega+10 \Omega=15 \Omega
\]
OR
(ii) \(\displaystyle \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{6 \mathrm{~V}}{(0.1+0.2+0.3+0.4+0.5) \Omega}=\frac{6 \mathrm{~V}}{1.5 \Omega}=4 \cdot 0 \mathrm{~A}\)
same current flows when resistors are connected in series.
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CBSE Class 10 Science past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.