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Physics · 2026 · 5 marks
CBSE 2026 · Region 3 · Set 1 · Q32
(a) (i) A light bulb and an open coil inductor are connected in series across an ac source of variable frequency. How will the glow of the bulb be…
(i)
A light bulb and an open coil inductor are connected in series across an ac source of variable frequency. How will the glow of the bulb be affected when :
(I)
an iron bar is inserted inside the coil, and
(II)
the frequency of the source is decreased ? Justify your answers. Assume that in each above case other factors remain unchanged.
(ii)
An ac voltage $\displaystyle \mathrm{V}=280 \sin (100 \pi \mathrm{t})$ volt is connected across a series LCR circuit in which $\displaystyle \mathrm{R}=400 \Omega, \mathrm{~L}=\frac{5}{\pi} \mathrm{H}$ and $\displaystyle \mathrm{C}=\frac{50}{\pi} \mu \mathrm{~F}$. Taking $\displaystyle \sqrt{2}=1 \cdot 4$, calculate :
(I)
impedance of the circuit.
(II)
rms value of current that flows in the circuit.
(III)
power factor of the circuit.
(i)
State Lenz's law and explain that it follows the law of conservation of energy.
(ii)
Write the dimensional formula for self-inductance. The current in a coil changes from $\displaystyle 8.0$ A to $\displaystyle 2.0$ A in $\displaystyle 0.6$ s. If an average emf induced in the coil is $\displaystyle 50$ V, calculate the self-inductance of the coil.
Marking-scheme solution
(i)
(I)
Brightness of the bulb decreases.
As the iron bar is inserted, the inductive reactance of the coil increases. A larger fraction of the applied AC voltage appears across the inductor, leaving less voltage across the bulb. (Alternatively: $\displaystyle L^{\prime}=\mu_{r} L$, L increases, $\displaystyle \mathrm{X}_{\mathrm{L}}=\omega L$ increases, current decreases.)
(II)
Brightness of the bulb increases.
As the frequency of the source decreases, the inductive reactance of the coil decreases. Less fraction of the applied AC voltage appears across the inductor, hence a higher voltage appears across the bulb. (Alternatively: $\displaystyle X_{L}=\omega L$, $\displaystyle \omega$ decreases, $\displaystyle X_{L}$ decreases, current increases.)
(ii)
$\displaystyle \mathrm{V}=280 \sin (100 \pi t)$
$\displaystyle \mathrm{X}_{\mathrm{L}}=100 \pi \times \dfrac{5}{\pi}=500\ \Omega$
$\displaystyle \mathrm{X}_{\mathrm{C}}=\dfrac{10^{6}}{100 \pi \times \dfrac{50}{\pi}}=200\ \Omega$
(I)
$\displaystyle Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}=\sqrt{(400)^{2}+(300)^{2}}=500\ \Omega$
(II)
$\displaystyle \mathrm{I}_{\text{rms}}=\left(\dfrac{V_{0}}{Z \sqrt{2}}\right)=\dfrac{280}{500 \times 1.4}=0.4 \mathrm{~A}$
(III)
Power factor $\displaystyle =\cos \phi=\mathrm{R} / \mathrm{Z}=400 / 500=0.8$
(i)
The polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it.
Consider the north pole of a magnet being moved towards a coil. The magnetic flux through the coil increases, hence current is induced in the counter-clockwise direction, making that side of the coil behave as a north pole and causing repulsion.
If this were not the case, then bringing a magnet towards a coil would construct a perpetual motion machine by a suitable arrangement. Therefore the production of an induced emf in a direction which opposes the change in magnetic flux follows from the law of conservation of energy.
(ii)
$\displaystyle \left[\mathrm{M} L^{2} T^{-2} A^{-2}\right]$
$\displaystyle \varepsilon=-L \dfrac{d I}{d t}$
$\displaystyle \mathrm{L}=\dfrac{-(50) \times 0.6}{(2-8)}$
$\displaystyle =5 \mathrm{~H}$
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CBSE Class 12 Physics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.