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Physics · 2026 · 3 marks
CBSE 2026 · Region 1 · Set 2 · Q26
A 12.0 μ F capacitor is charged to a potential difference of 150 V. The terminals of the charged capacitor are then connected to those of an…
A $\displaystyle 12.0 \mu \mathrm{~F}$ capacitor is charged to a potential difference of $\displaystyle 150$ V. The terminals of the charged capacitor are then connected to those of an uncharged $\displaystyle 6.0 ~ \mu \mathrm{~F}$ capacitor. Calculate final potential difference across and charge on, each capacitor.
Marking-scheme solution
Initial charge on $\displaystyle 12 \mu \mathrm{~F}$ capacitor is $\displaystyle \mathrm{Q}_{\mathrm{i}}=\mathrm{C}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}$
$\displaystyle \mathrm{Q}_{\mathrm{i}}=12 \mu \mathrm{F} \times 150 \mathrm{~V}$
$\displaystyle \mathrm{Q}_{\mathrm{i}}=1800 \mu \mathrm{C}$
When $\displaystyle 6 \mu \mathrm{~F}$ capacitor is connected with $\displaystyle 12 \mu \mathrm{~F}$ capacitor then common potential of capacitor
$\displaystyle \mathrm{V}_{\mathrm{f}}=\dfrac{\mathrm{C}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
$\displaystyle \mathrm{V}_{\mathrm{f}}=\dfrac{1800 \mu \mathrm{C}}{(12+6) \mu \mathrm{F}}$
$\displaystyle \mathrm{V}_{\mathrm{f}}=100 \mathrm{~V}$
Final voltage on each capacitor is $\displaystyle 100$ V
Charge on $\displaystyle 12 \mu \mathrm{~F}$ capacitor is $\displaystyle \mathrm{Q}_{1}=\mathrm{C}_{1} \mathrm{V}_{\mathrm{f}}$
$\displaystyle \mathrm{Q}_{1}=12 \mu \mathrm{F} \times 100 \mathrm{~V}$
$\displaystyle \mathrm{Q}_{1}=1200 \mu \mathrm{C}$
Charge of $\displaystyle 6 \mu \mathrm{~F}$ capacitor $\displaystyle \mathrm{Q}_{2}=\mathrm{C}_{2} \mathrm{V}_{\mathrm{f}}$
$\displaystyle \mathrm{Q}_{2}=6 \mu \mathrm{F} \times 100 \mathrm{~V}$
$\displaystyle \mathrm{Q}_{2}=600 \mu \mathrm{C}$
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CBSE Class 12 Physics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.