CBSE 2025 · Region 2 · Set 3 · Q24 · 3 marks
(a)ac voltage of frequency $\displaystyle \omega$ is applied across a series LCR circuit. Draw the phasor diagram and obtain the impedance of the circuit.(b)Discuss'resonance'in a series LCR circuit and write the expression for resonant frequency.
(a)
ac voltage of frequency $\displaystyle \omega$ is applied across a series LCR circuit. Draw the phasor diagram and obtain the impedance of the circuit.
(b)
Discuss'resonance'in a series LCR circuit and write the expression for resonant frequency.
Marking-scheme solution
(a)
$\displaystyle \vec{V}_{L}+\vec{V}_{R}+\vec{V}_{C}=\vec{V}$
$\displaystyle v_{m}^{2}=v_{R m}^{2}+\left(v_{C m}-v_{L m}\right)^{2}$
$\displaystyle =\left(i_{m} R\right)^{2}+\left(i_{m} X_{C}-i_{m} X_{L}\right)^{2}$
$\displaystyle v_{m}^{2}=i_{m}^{2}\left[R^{2}+\left(X_{C}-X_{L}\right)^{2}\right]$
$\displaystyle i_{m}=\frac{v_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}}$
$\displaystyle \Rightarrow Z=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}$
(b)
When $\displaystyle X_{L}=X_{C}$, the impedance of the series LCR circuit is minimum and the current flowing through it is maximum.
$\displaystyle \omega_{0}=\frac{1}{\sqrt{L C}}$
Alternating CurrentAC Voltage Applied to a Series LCR CircuitApplyshort_answermedium
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CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.