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CBSE Class 12 Physics 2023 Outside Delhi Set 2 Paper

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Question : 11 of 13
Marks: +1, -0
SECTION - C
A series RLR L circuit with R=10 ΩR=10\,\Omega and L=(100π) mHL=\left(\frac{100}{\pi}\right)\,\text{mH} is connected to an ac source of voltage V=141 sin⁡V=141\,\sin (100πt)(100\pi t), where VV is in volts and tt is in seconds. Calculate
(a) impedance of the circuit
(b) phase angle, and
(c) voltage drop across the inductor
Solution:     👈: Video Solution
(a) Given, R=10 ΩR=10\,\Omega
L=(100π) mHL=\left(\frac{100}{\pi}\right)\,\text{mH}
V=141sin⁡(100πt)V=141\sin(100\pi t)
Impedance=Z=R2+(ωL)2\text{Impedance}=Z=\sqrt{R^2+(\omega L)^2}
Or, Z=102+(100π×100π×10−3)2Z=\sqrt{10^2+\left(100\pi \times \frac{100}{\pi} \times 10^{-3}\right)^2}
Or, Z=102+102Z=\sqrt{10^2+10^2}
∴ Z=102 ΩZ=10\sqrt{2}\,\Omega
(b) Phase angle=tan⁡−1ωLR\text{Phase angle}=\tan^{-1}\frac{\omega L}{R}
Or, Phase angle =tan⁡−1(100π×100π×10−310)=\tan^{-1}\left(\frac{100\pi \times \frac{100}{\pi} \times 10^{-3}}{10}\right)
Or,     \;\; Phase angle =tan⁡−11=\tan^{-1}1
∴\therefore Phase angle =45∘=45^{\circ}
(c) V0=141 VV_0=141\,\text{V}
XL=ωLX_L=\omega L
=100π×100π×10−3=100\pi \times \frac{100}{\pi} \times 10^{-3}
=10 Ω=10\,\Omega
I0=V0Z=141102I_0=\frac{V_0}{Z}=\frac{141}{10\sqrt{2}}
Voltage drop across inductor =VL=I0XL=V_L=I_0 X_L
Or, VL=141102×10\text{Or, } V_L=\frac{141}{10\sqrt{2}} \times 10
∴VL=100 V\therefore V_L=100\,\text{V}
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