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CBSE Class 12 Physics 2023 Outside Delhi Set 1 Paper

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Question : 28 of 35
Marks: +1, -0
A series CR circuit with R=200 ΩR=200\,\Omega and C=(50π)C=\left(\frac{50}{\pi}\right) μ F\mu\,\mathrm{F} is connected across an ac source of peak voltage ε0=100 V\varepsilon_0=100\,\mathrm{V} and frequency v=50 Hzv=50\,\mathrm{Hz}. Calculate (a) impedance of the circuit (Z),(b)(Z),(b) phase angle (ϕ)(\phi), and (c) voltage across the resistor.
Solution:     👈: Video Solution
(a) Given, R=200 Ω, C=50π μFR=200\,\Omega,\,C=\frac{50}{\pi}\,\mu\mathrm{F}
Impedance: Z=R2+(1ωC)2Z=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}
Or,
Z=(200)2+(12π×50×50π×10−6)2Z=\sqrt{(200)^2+\left(\frac{1}{2\pi\times50\times\frac{50}{\pi}\times10^{-6}}\right)^2}
Or,
Z=(200)2+(200)2Z=\sqrt{(200)^2+(200)^2}
∴Z=2002\therefore Z=200\sqrt{2}
(b) Phase angle:
θ=tan⁡−11ωCR\theta=\tan^{-1}\frac{1}{\omega C R}
or, θ=tan⁡−1(12π×50×50π×10−6×200)\theta=\tan^{-1}\left(\frac{1}{2\pi\times50\times\frac{50}{\pi}\times10^{-6}\times200}\right)
or, θ=tan⁡−11\theta=\tan^{-1}1
∴ θ=45∘\theta=45^{\circ}
(c) Voltage across resistor:
IRMS=εRMSZI_{\mathrm{RMS}}=\frac{\varepsilon_{\mathrm{RMS}}}{Z}
Or,\text{Or,} IRMS=ε02ZI_{\mathrm{RMS}}=\frac{\frac{\varepsilon_0}{\sqrt{2}}}{Z}
Or, IRMS=10022002I_{\mathrm{RMS}}=\frac{\frac{100}{\sqrt{2}}}{200\sqrt{2}}
∴IRMS=14 A\therefore I_{\mathrm{RMS}}=\frac{1}{4}\,\mathrm{A}
VR=IRMS×RV_R=I_{\mathrm{RMS}}\times R
Or, VR=14×200V_R=\frac{1}{4}\times200
∴VR=50 V\therefore V_R=50\,\mathrm{V}
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