CBSE 2023 · Region 5 · Set 2 · Q27 · 3 marks
A current of $\displaystyle 1$ A flows through a coil when it is connected across a DC battery of $\displaystyle 100$ V . If DC battery is replaced by an AC source of $\displaystyle 100$ V and angular frequency $\displaystyle 100 \mathrm{rad} \mathrm{s}^{-1}$, the current reduces to $\displaystyle 0.5$ A . Find(i)impedance of the circuit.(ii)self-inductance of coil.(iii)phase difference between the voltage and the current.
A current of $\displaystyle 1$ A flows through a coil when it is connected across a DC battery of $\displaystyle 100$ V . If DC battery is replaced by an AC source of $\displaystyle 100$ V and angular frequency $\displaystyle 100 \mathrm{rad} \mathrm{s}^{-1}$, the current reduces to $\displaystyle 0.5$ A . Find
(i)
impedance of the circuit.
(ii)
self-inductance of coil.
(iii)
phase difference between the voltage and the current.
Marking-scheme solution
(i)
Impedance $\displaystyle Z = \dfrac{V_{rms}}{I_{rms}}$
$\displaystyle Z = \dfrac{100}{0.5}$
$\displaystyle Z = 200\ \Omega$
(ii)
$\displaystyle Z^2 = X_L^2 + R^2$
Resistance $\displaystyle R = \dfrac{V_{rms}}{I_{rms}} = \dfrac{100}{1} = 100\ \Omega$
$\displaystyle (X_L)^2 = Z^2 - R^2$
$\displaystyle = (200)^2 - (100)^2$
$\displaystyle X_L = 100\sqrt{3}$
$\displaystyle \omega L = 100\sqrt{3}$
$\displaystyle 100L = 100\sqrt{3}$
$\displaystyle \Rightarrow L = \sqrt{3}\ H$
(iii)
Phase difference
$\displaystyle \tan\phi = \dfrac{X_L}{R}$
$\displaystyle = \dfrac{100\sqrt{3}}{100} = \sqrt{3}$
$\displaystyle \Rightarrow \phi = 60^{\circ}$
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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.