CBSE 2023 · Region 4 · Set 1 · Q28 · 3 marks
A resistor of $\displaystyle 30 \Omega$ and a capacitor of $\displaystyle \frac{250}{\pi} \mu \mathrm{~F}$ are connected in series to a $\displaystyle 200 \mathrm{~V}, 50 \mathrm{~Hz}$ ac source. Calculate (i) the current in the circuit, and (ii) voltage drops across the resistor and the capacitor. (iii) Is the algebraic sum of these voltages more than the source voltage? If yes, solve the paradox.A series LCR circuit with R $\displaystyle =20 \Omega, \mathrm{~L}=2 \mathrm{H}$ and $\displaystyle \mathrm{C}=50 \mu \mathrm{~F}$ is connected to a $\displaystyle 200$ volts ac source of variable frequency. What is(i)the amplitude of the current, and (ii) the average power transferred to the circuit in one complete cycle, at resonance ?(iii)Calculate the potential drop across the capacitor.
A resistor of $\displaystyle 30 \Omega$ and a capacitor of $\displaystyle \frac{250}{\pi} \mu \mathrm{~F}$ are connected in series to a $\displaystyle 200 \mathrm{~V}, 50 \mathrm{~Hz}$ ac source. Calculate (i) the current in the circuit, and (ii) voltage drops across the resistor and the capacitor. (iii) Is the algebraic sum of these voltages more than the source voltage? If yes, solve the paradox.
A series LCR circuit with R $\displaystyle =20 \Omega, \mathrm{~L}=2 \mathrm{H}$ and $\displaystyle \mathrm{C}=50 \mu \mathrm{~F}$ is connected to a $\displaystyle 200$ volts ac source of variable frequency. What is
(i)
the amplitude of the current, and (ii) the average power transferred to the circuit in one complete cycle, at resonance ?
(iii)
Calculate the potential drop across the capacitor.
Marking-scheme solution
(a) \[\begin{aligned}
& \because X_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}} \\
& \omega=2 \pi v=100 \pi \\
& X_{\mathrm{C}}=\frac{1}{100 \pi \times 250 / \pi \times 10^{-6}} \\
& \quad=40 \Omega
\end{aligned}
\]
Impedance of the circuit
\[\begin{aligned}
Z & =\sqrt{X_{\mathrm{C}}^{2}+\mathrm{R}^{2}} \\
& =\sqrt{(40)^{2}+(30)^{2}}=50 \Omega
\end{aligned}
\]
(i) Current in the circuit
\[I_{r m s}=\frac{\mathrm{V}_{r m s}}{Z}=\frac{200}{50}=4 \mathrm{~A}
\]
(ii) Voltage drops across the Capacitor,
\[\mathrm{V}_{\mathrm{C}}=I_{r m s} X_{\mathrm{C}}=4 \times 40=160 \mathrm{~V}
\]
Voltage drops across the Resistor,
\[\mathrm{V}_{\mathrm{R}}=I_{\text {rms }} \times \mathrm{R}=4 \times 30=120 \mathrm{~V}
\]
(iii)
The algebraic sum of the two voltages \(\displaystyle \mathrm{V}_{\mathrm{R}}\) and \(\displaystyle \mathrm{V}_{\mathrm{C}}\) is $\displaystyle 280$ V , which
Alternating CurrentAC Voltage Applied to a Series LCR CircuitApplyshort_answermedium
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CBSE Class 12 Physics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.