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Physics · 2025 · 3 marks
CBSE 2025 · Region 1 · Set 3 · Q24
An ac source of voltage ν=ν_m sin ω t is connected to a series combination of LCR circuit. Draw the phasor diagram. Using it obtain an expression for…
An ac source of voltage $\displaystyle \nu=\nu_{\mathrm{m}} \sin \omega \mathrm{t}$ is connected to a series combination of LCR circuit. Draw the phasor diagram. Using it obtain an expression for the impedance of the circuit and the phase difference between applied voltage and the current.
Marking-scheme solution
$\displaystyle V_{R m}=i_{m} R, V_{C m}=i_{m} X_{C}, V_{L m}=i_{m} X_{L}$
From phasor diagram:
$\displaystyle V_{m}^{2}=V_{R m}^{2}+\left(V_{C m}-V_{L m}\right)^{2}$
$\displaystyle V_{m}^{2}=\left(i_{m} R\right)^{2}+\left(i_{m} X_{C}-i_{m} X_{L}\right)^{2}$
$\displaystyle =i_{m}^{2}\left[R^{2}+\left(X_{C}-X_{L}\right)^{2}\right]$
Or $\displaystyle i_{m}=\frac{V_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}}$
$\displaystyle \therefore i_{m}=\frac{V_{m}}{Z}$
$\displaystyle \therefore Z=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}$
From phasor diagram:
$\displaystyle \tan \theta=\frac{V_{C m}-V_{L m}}{V_{R m}}$
$\displaystyle =\frac{X_{C}-X_{L}}{R}$
$\displaystyle \therefore \theta=\tan ^{-1}\left(\frac{X_{C}-X_{L}}{R}\right)$
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CBSE Class 12 Physics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.