CBSE 2024 · Region 1 · Set 1 · Q32 · 5 marks
(i)A resistor and a capacitor are connected in series to an ac source $\displaystyle \mathrm{v}=\mathrm{v}_{\mathrm{m}} \sin \omega \mathrm{t}$. Derive an expression for the impedance of the circuit.(ii)When does an inductor act as a conductor in a circuit ? Give reason for it.(iii)An electric lamp is designed to operate at $\displaystyle 110$ V dc and $\displaystyle 11$ A current. If the lamp is operated on $\displaystyle 220 \mathrm{~V}, 50 \mathrm{~Hz}$ ac source with a coil in series, then find the inductance of the coil.(i)Draw a labelled diagram of a step-up transformer and describe its working principle. Explain any three causes for energy losses in a real transformer.(ii)A step-up transformer converts a low voltage into high voltage. Does it violate the principle of conservation of energy? Explain.(iii)A step-up transformer has $\displaystyle 200$ and $\displaystyle 3000$ turns in its primary and secondary coils respectively. The input voltage given to the primary coil is $\displaystyle 90$ V . Calculate :(1)The output voltage across the secondary coil(2)The current in the primary coil if the current in the secondary coil is $\displaystyle 2 \cdot 0 \mathrm{~A}$.
(i)
A resistor and a capacitor are connected in series to an ac source $\displaystyle \mathrm{v}=\mathrm{v}_{\mathrm{m}} \sin \omega \mathrm{t}$. Derive an expression for the impedance of the circuit.
(ii)
When does an inductor act as a conductor in a circuit ? Give reason for it.
(iii)
An electric lamp is designed to operate at $\displaystyle 110$ V dc and $\displaystyle 11$ A current. If the lamp is operated on $\displaystyle 220 \mathrm{~V}, 50 \mathrm{~Hz}$ ac source with a coil in series, then find the inductance of the coil.
(i)
Draw a labelled diagram of a step-up transformer and describe its working principle. Explain any three causes for energy losses in a real transformer.
(ii)
A step-up transformer converts a low voltage into high voltage. Does it violate the principle of conservation of energy? Explain.
(iii)
A step-up transformer has $\displaystyle 200$ and $\displaystyle 3000$ turns in its primary and secondary coils respectively. The input voltage given to the primary coil is $\displaystyle 90$ V . Calculate :
(1)
The output voltage across the secondary coil
(2)
The current in the primary coil if the current in the secondary coil is $\displaystyle 2 \cdot 0 \mathrm{~A}$.
Marking-scheme solution
(a)
$$V_C + V_R = V$$
$$v_m^2 = v_{rm}^2 + v_{cm}^2$$
$$v_{rm} = i_m R$$
$$v_{cm} = i_m X_c$$
$$v_m^2 = (i_m R)^2 + (i_m X_c)^2$$
$$= i_m^2\left[R^2 + X_c^2\right]$$
$$\Rightarrow i_m = \frac{v_m}{\sqrt{R^2 + X_c^2}}$$
$$\Rightarrow \text{Impedance } Z = \sqrt{R^2 + X_c^2}$$
(ii)
For direct current (dc), an inductor behaves as a conductor.
As $\displaystyle X_L = \omega L = 2\pi \nu L$
For dc $\displaystyle \nu = 0 \Rightarrow X_L = 0$
Alternatively:-
Induced emf $\displaystyle (\varepsilon) = -\dfrac{L\,dI}{dt}$
For dc; $\displaystyle dI = 0 \Rightarrow \varepsilon = 0$
(iii)
$\displaystyle R = \dfrac{110}{11} = 10\,\Omega$
$$i_{rms} = \frac{v_{rms}}{\sqrt{R^2 + X_L^2}} = \frac{220}{\sqrt{100 + X_L^2}}$$
$$11 = \frac{220}{\sqrt{100 + X_L^2}}$$
$$\sqrt{100 + X_L^2} = \frac{220}{11} = 20\,\Omega$$
Squaring both sides:
$$\Rightarrow 100 + X_L^2 = 400$$
$$\Rightarrow X_L^2 = 300 \Rightarrow X_L = 10\sqrt{3}\,\Omega$$
$$X_L = 2\pi f L \Rightarrow 10\sqrt{3} = 2\pi \times 50 \times L$$
$$L = \frac{\sqrt{3}}{10\pi}\,H$$
(b)
The working principle of transformer is mutual induction.
When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it.
Causes of energy losses
(a)
Flux leakage
(b)
Resistance of the windings
Eddy currents
(d)
Hysteresis
(ii)
No
Current changes correspondingly. So, the input power is equal to the output power.
(iii)
($\displaystyle 1$)
$$\frac{V_s}{V_P} = \frac{N_s}{N_P}$$
$$V_s = \frac{N_s}{N_P} \times V_P = \frac{3000}{200} \times 90$$
$$V_s = 1350\,V$$
(2)
$$\frac{I_P}{I_s} = \frac{N_s}{N_P}$$
$$I_P = \frac{3000}{200} \times 2 = 30\ \text{A}$$
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.