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Science · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 3 · Q39
(a) Study the given electric circuit in which 2 A electric current is flowing between points X and Y. (i) Using the battery, key, voltmeter and…
Study the given electric circuit in which $\displaystyle 2$ A electric current is flowing between points X and Y.
(i)
Using the battery, key, voltmeter and ammeter in this given electric circuit, redraw a circuit diagram in which $\displaystyle 2 \Omega$ and $\displaystyle 4 \Omega$ resistors are connected between X and Y in parallel combination.
(ii)
Circuit drawn by you, in which resistors are connected in parallel combination, calculate the electric current flowing through $\displaystyle 4 \Omega$ resistor.
(i)
Two lamps 'A' and 'B' of rating $\displaystyle 50$ W; $\displaystyle 220$ V and $\displaystyle 100$ W, $\displaystyle 220$ V are connected in series combination. Find out the ratio of the resistances $\displaystyle \left(\mathrm{R}_{\mathrm{A}}: \mathrm{R}_{\mathrm{B}}\right)$ of these lamps.
(ii)
Derive the expression for the equivalent resistance of three resistors $\displaystyle \mathrm{R}_{1}, \mathrm{R}_{2}$ and $\displaystyle \mathrm{R}_{3}$ connected in parallel combination.
Marking-scheme solution
(i) 
Diagram- $\displaystyle 1$ mark
Diagram- $\displaystyle 1$ mark Labelling - $\displaystyle 2$ marks Labelling - $\displaystyle 2$ marks
Diagram- $\displaystyle 1$ mark
Diagram- $\displaystyle 1$ mark Labelling - $\displaystyle 2$ marks Labelling - $\displaystyle 2$ marks(ii) Potential Difference across the battery for \(\displaystyle 4 \Omega\) and \(\displaystyle 2 \Omega\) resistors connected in series,
\[\begin{aligned}
& \mathrm{V}=\mathrm{IR} \\
& \mathrm{~V}=2 \times 6=12 \mathrm{~V}
\end{aligned}
\] Current through \(\displaystyle 4 \Omega\) resistor in circuit shown in (i),
\[\begin{aligned}
& \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \\
& \mathrm{I}=\frac{12}{4} \\
& \mathrm{I}=3 \mathrm{~A}
\end{aligned}
\] OR
(b)
(i) \(\displaystyle \mathrm{R}_{\mathrm{A}}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{\mathrm{A}}}\)
\[\mathrm{R}_{\mathrm{A}}=\frac{220 \times 220}{50}
\] \(\displaystyle \mathrm{R}_{\mathrm{A}}=968 \Omega\)
\[R_{B}=\frac{V^{2}}{P_{B}}
\] \(\displaystyle \mathrm{R}_{\mathrm{B}}=\frac{220 \times 220}{100} \mathrm{R}_{\mathrm{B}}=484 \Omega \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{968}{484} \mathrm{R}_{\mathrm{A}}: \mathrm{R}_{\mathrm{B}}=2: 1\)
(ii)
As in parallel combination, Potential difference applied across each resistor is same. Current flowing through each resistor \(\displaystyle \mathrm{R}_{1}, \mathrm{R}_{2}\) and \(\displaystyle \mathrm{R}_{3}\) is
\[\mathrm{I}_{1}=\frac{\mathrm{V}}{\mathrm{R}_{1}} ; \mathrm{I}_{2}=\frac{\mathrm{V}}{\mathrm{R}_{2}} ; \mathrm{I}_{3}=\frac{\mathrm{V}}{\mathrm{R}_{3}}
\]
If resistance of parallel combination is \(\displaystyle \mathrm{R}_{\mathrm{P}}\) then current drawn from battery is
\[I=\frac{V}{R_{p}}
\]
Current \(\displaystyle \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}\)
\[\frac{V}{R_{P}}=\frac{V}{R_{1}}+\frac{V}{R_{2}}+\frac{V}{R_{3}}
\]
\[\frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}
\]
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CBSE Class 10 Science past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.