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Physics · 2023 · 3 marks
CBSE 2023 · Region 1 · Set 3 · Q26
An ac voltage v = v_m sin ω t ( v_m =310 V. and. f =50 Hz ) is connected to a pure capacitor of capacitance 15 μ F. Calculate (i) the reactance of…
An ac voltage $\displaystyle \mathrm{v}=\mathrm{v}_{\mathrm{m}} \sin \omega \mathrm{t}\left(\mathrm{v}_{\mathrm{m}}=310 \mathrm{~V}\right.$ and $\displaystyle \left.\mathrm{f}=50 \mathrm{~Hz}\right)$ is connected to a pure capacitor of capacitance $\displaystyle 15 \mu \mathrm{~F}$. Calculate (i) the reactance of the capacitor, and (ii) the amplitude of the current. Write the expression of current through the capacitor as a function of time.
Marking-scheme solution
\[\begin{array}{ll}
\text { (a) Calculation of reactance of capacitor } & 1 \\
\text { (b) Calculation of amplitude of current } & 1 \\
\text { Writing expression of current } & 1
\end{array}
\]
\(\displaystyle \mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t} \quad \mathrm{C}=15 \times 10^{-6} F \mathrm{V}=310 \sin 100 \pi \mathrm{t}\)
(i)
\(\displaystyle \mathrm{X}_{\mathrm{c}}=\frac{1}{\omega \mathrm{C}}\)
\[\begin{aligned}
& =\frac{1}{100 \pi \times 15 \times 10^{-6}} \\
& =\frac{10^{4}}{15 \pi}=212 \Omega
\end{aligned}
\]
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CBSE Class 12 Physics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.