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Science · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 3 · Q39 (b)
(i) Two lamps 'A' and 'B' of rating 50 W; 220 V and 100 W, 220 V are connected in series combination. Find out the ratio of the resistances (R_A: R_B…
(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 $\displaystyle \left(R_{\mathrm{A}}: R_{\mathrm{B}}\right)$ of these lamps.
(ii)
Derive the expression for the equivalent resistance of three resistors $\displaystyle R_{1}, R_{2}$ and $\displaystyle R_{3}$ connected in parallel combination.
Marking-scheme solution
(i)
$\displaystyle \mathrm{R}_{\mathrm{A}}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{\mathrm{A}}}$
$\displaystyle \mathrm{R}_{\mathrm{A}}=\frac{220 \times 220}{50}$
$\displaystyle \mathrm{R}_{\mathrm{A}}=968 \Omega$
$\displaystyle \mathrm{R}_{\mathrm{B}}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{\mathrm{B}}}$
$\displaystyle \mathrm{R}_{\mathrm{B}}=\frac{220 \times 220}{100}$
$\displaystyle \mathrm{R}_{\mathrm{B}}=484 \Omega$
$\displaystyle \frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{968}{484}$
$\displaystyle \mathrm{R}_{\mathrm{A}}: \mathrm{R}_{\mathrm{B}}=2: 1$
(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
$\displaystyle \mathrm{I}_{1}=\frac{\mathrm{V}}{\mathrm{R}_{1}} ; \mathrm{I}_{2}=\frac{\mathrm{V}}{\mathrm{R}_{2}} ; \mathrm{I}_{3}=\frac{\mathrm{V}}{\mathrm{R}_{3}}$
If resistance of parallel combination is $\displaystyle \mathrm{R}_{\mathrm{P}}$ then current drawn from battery is
$\displaystyle \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{P}}}$
Current $\displaystyle \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}$
$\displaystyle \frac{\mathrm{V}}{\mathrm{R}_{\mathrm{P}}}=\frac{\mathrm{V}}{\mathrm{R}_{1}}+\frac{\mathrm{V}}{\mathrm{R}_{2}}+\frac{\mathrm{V}}{\mathrm{R}_{3}}$
$\displaystyle \frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}+\frac{1}{\mathrm{R}_{3}}$
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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.