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Physics · 2026 · 5 marks
CBSE 2026 · Region 4 · Set 1 · Q32
(a) (i) Define the terms (I) resonant frequency, and (II) power factor of a series LCR circuit. For what value of the power factor will the power…
(i)
Define the terms (I) resonant frequency, and (II) power factor of a series LCR circuit. For what value of the power factor will the power dissipated in the circuit be maximum ?
(ii)
An inductor of $\displaystyle \frac{5}{\pi} \mathrm{H}$, a capacitor of $\displaystyle \frac{50}{\pi} \mu \mathrm{~F}$ and a resistor of $\displaystyle 400 \Omega$ are connected in series across an ac voltage $\displaystyle \mathrm{v}=140 \sin (100 \pi \mathrm{t}) \mathrm{V}$. Calculate :
(I)
impedance of the circuit, and
(II)
rms value of current that flows in the circuit. (Take $\displaystyle \sqrt{2}=1 \cdot 4$ )
(i)
Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary voltage to primary voltage in terms of number of turns in the two coils.
(ii)
The number of turns in the primary and the secondary coil of an ideal transformer are $\displaystyle 100$ and $\displaystyle 5000$ respectively. If $\displaystyle 3 \cdot 3$ kW power is supplied to the transformer at $\displaystyle 220$ V, find
(I)
current in the primary coil, and (II) output voltage.
Marking-scheme solution
(i)
(I)
Resonant frequency — for a series LCR circuit, the frequency at which the current amplitude is maximum. (Alternatively: the frequency at which the impedance is minimum, $\displaystyle Z=R$; or the frequency at which $\displaystyle \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$.)
(II)
Power factor — it is the ratio of resistance to impedance of the series LCR circuit. (Alternatively: $\displaystyle \cos \phi=\dfrac{R}{Z}$, where $\displaystyle \phi$ is the angle between the voltage and the current; or $\displaystyle \cos \phi=\dfrac{\mathrm{V}_{\mathrm{R}}}{\mathrm{V}}$.)
$\displaystyle P=\mathrm{V}_{\text{eff}} \mathrm{I}_{\text{eff}} \cos \phi$; P is maximum when $\displaystyle \cos \phi$ is unity.
(ii)
(I)
$\displaystyle Z=\sqrt{R^{2}+\left(\omega \mathrm{L}-\dfrac{1}{\omega c}\right)^{2}}=\sqrt{(400)^{2}+\left(100 \pi \times \dfrac{5}{\pi}-\dfrac{1 \times \pi}{100 \pi \times 50 \times 10^{-6}}\right)^{2}}$
$\displaystyle =\sqrt{160000+90000}=\sqrt{250000}=500\ \Omega$
(II)
$\displaystyle \mathrm{I}_{\text{rms}}=\dfrac{V_{0}}{Z \sqrt{2}}=\dfrac{140}{500 \times 1.4}=0.2 \mathrm{~A}$
(i)
The induced emf or voltage across the secondary with $\displaystyle \mathrm{N}_{\mathrm{s}}$ turns is $\displaystyle \varepsilon_{s}=-\mathrm{N}_{\mathrm{s}} \dfrac{d \phi}{d t}$
The alternating flux $\displaystyle \phi$ also induces an emf, called back emf: $\displaystyle \varepsilon_{\mathrm{P}}=-\mathrm{N}_{\mathrm{P}} \dfrac{d \phi}{d t}$
$\displaystyle \varepsilon_{\mathrm{P}}=v_{\mathrm{P}}$ (the primary coil has zero resistance) and $\displaystyle \varepsilon_{s}=v_{s}$ (the secondary coil is an open circuit)
$\displaystyle v_{s}=-\mathrm{N}_{\mathrm{s}} \dfrac{d \phi}{d t} \quad \ldots(1)$
$\displaystyle v_{\mathrm{P}}=-\mathrm{N}_{\mathrm{P}} \dfrac{d \phi}{d t} \quad \ldots(2)$
Dividing ($\displaystyle 1$) by ($\displaystyle 2$): $\displaystyle \dfrac{v_{s}}{v_{\mathrm{P}}}=\dfrac{\mathrm{N}_{s}}{\mathrm{~N}_{\mathrm{P}}}$
(ii)
(I)
$\displaystyle P=\mathrm{V}_{\mathrm{P}} \mathrm{I}_{\mathrm{P}}$: $\displaystyle 3.3 \times 10^{3}=220 \times \mathrm{I}_{\mathrm{P}}$
$\displaystyle \mathrm{I}_{\mathrm{P}}=\dfrac{3.3 \times 10^{3}}{220}=15 \mathrm{~A}$
(II)
$\displaystyle \dfrac{v_{s}}{v_{\mathrm{P}}}=\dfrac{\mathrm{N}_{s}}{\mathrm{~N}_{\mathrm{P}}}$: $\displaystyle \dfrac{220}{v_{s}}=\dfrac{100}{5000}$
$\displaystyle v_{s}=11000 \mathrm{~V}$
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