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Physics · 2023 · 3 marks
CBSE 2023 · Region 2 · Set 2 · Q26
A series RL circuit with R =10 Ω and L =(100/(π)) mH is connected to an ac source of voltage V=141 sin (100 π t), where V is in volts and t is in…
A series RL circuit with $\displaystyle \mathrm{R}=10 \Omega$ and $\displaystyle \mathrm{L}=\left(\frac{100}{\pi}\right) \mathrm{mH}$ is connected to an ac source of voltage $\displaystyle V=141 \sin (100 \pi t)$, where $\displaystyle V$ is in volts and $\displaystyle t$ is in seconds. Calculate
(a)
impedence of the circuit
(b)
phase angle, and
(c)
voltage drop across the inductor
Marking-scheme solution
(a)
$\displaystyle Z = \sqrt{R^2 + X_L^2}$ , $\displaystyle X_L = \omega L$
$\displaystyle = \sqrt{100 + \left(100\pi \times \dfrac{100}{\pi} \times 10^{-3}\right)^2}$
$\displaystyle = 10\sqrt{2}\ \Omega$
(b)
$\displaystyle \tan\phi = \dfrac{X_L}{R}$
$\displaystyle = 1$
$\displaystyle \therefore\ \phi = \dfrac{\pi}{4}\ or\ 45^0$
(c)
$\displaystyle V_{rms} = I_{rms} X_L$
$\displaystyle = \dfrac{141 \times 10}{\sqrt{2} \times 10\sqrt{2}}$
$\displaystyle = \dfrac{141}{2}\ \mathrm{V} = 70.5\ \mathrm{V}$
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CBSE Class 12 Physics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.