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Published on: 13/05/2022
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Take MCQ Physics Test1.
The 300 turn primary of a transformer has resistance 0.82 Ω and the resistance of its secondary of 1200 turns is 6.2 Ω. Find the voltage across the primary if the power output from the secondary at 1600V is 32 kW. Calculate the power losses in both coils when the transformer efficiency is 80%.
2.
A graph between the magnitude of the magnetic flux linked with a closed loop and time is given in the figure. Arrange the regions of the graph in ascending order of the magnitude of induced emf in the loop.

3.
A flexible metallic loop abcd in the shape of a square is kept in a magnetic field with its plane perpendicular to the field. The magnetic field is directed into the paper normally. Find the direction of the induced current when the square loop is crushed into an irregular shape as shown in the figure.

4.
In series LC circuit, the voltages across L and C are 180° out of phase. Is it correct? Explain.
5.
Draw graphs showing the distribution of charge in a capacitor and current through an inductor during LC oscillations with respect to time. Assume that the charge in the capacitor is maximum initially.
1.
\(N_z=300, N_s=1200, R_p=0.82 \Omega, R_s=6.2 \Omega, \)
\(\text {Output power }=32 \mathrm{~kW}, \mathrm{~V}_{\mathrm{s}}=1600 \mathrm{~V}, \eta=80 \% \)
\(V_s I_s=32000 \mathrm{~W} \)
\(1600 \times I_s=32000 \)
\(\therefore I_s=\frac{32000}{1600} \)
\(I_s=20 \mathrm{~A}\) ......(1)
∴ Power loss in secondary coil \(P =I_{\mathrm{s}}^2 \times R_{\mathrm{s}} \) ......(2)
\(=(20)^2 \times 6.2 \)
\(=400 \times 6.2=2480.0 \)
\(=2480 \mathrm{~W}=2.48 \mathrm{~kW}\)
∴ Power loss in secondary coil P = 2.48 kW ......(3)
Efficiency \( \eta =\frac{V_s I_s}{V_P I_P} \)
\(\therefore \frac{80}{100} =\frac{32 \times 10^3}{V_P I_P} \)
\(\therefore V_P I_p =\frac{32 \times 10^3 \times 100}{80} \)
\(=\frac{3200 \times 10^3}{80} \)
\(V_P I_P =40 \times 10^3 \mathrm{~W}\) ......(4)
We know that, \(\frac{N_p}{N_s}=\frac{V_P}{V_S}\) ......(5)
\(\frac{300}{1200}=\frac{V_p}{1600} \Rightarrow V_p =\frac{300 \times 1600}{1200} \)
\(=400 \mathrm{~V}\) ......(6)
From (4), VpIp = 40 x 103
400 Ip = 40 x 103
\(\therefore I_p =\frac{40 \times 10^3} { 400}={100A} \)
∴ Power loss in the primary coil = I2pRp
= (100)2 x 0.82
= 104 x 0.82 = 8200 W
∴ Power loss in the primary coil = 8.2 kW
2.
Magnitude of induced \(\mathrm{emf}|\mathrm{e}|=\left|-\frac{d \phi}{d t}\right|\)
\(\therefore c=\frac{d \phi}{d t}\)
(i) In the region ab,
\(c_1=\frac{d \phi_1}{d t}=\frac{4-0}{1-0}=\frac{4}{1}=4 \mathrm{~V}\)
(ii) In the region bc,
\(c_2=\frac{d \phi_2}{d t}=\frac{4-4}{3-1}=\frac{0}{2}=0 \mathrm{~V}\)
(iii) In the region cd,
\(e_3=\frac{d \phi_2}{d t}=\frac{4-2}{4-3}=\frac{2}{1}=2 \mathrm{~V}\)
(iv) In the region de,
Hence,
\(e_4=\frac{d \phi_4}{d t}=\frac{2-0}{7-4}=\frac{2}{3}=0.66 V \)
\(\mathrm{e}_2<\mathrm{e}_4<\mathrm{e}_3<\mathrm{e}_1\)
In the ascending order of the magnitude of induced emf
Region bc < Region de < Region cd < Region ab.
3.
The direction of induced current is, a → b, b → c, c → d and d → a
4.
In the inductor voltage leads current by 90°
In the capacitor current leads voltage by 90°
Hence in series LC circuit, the voltage across L and C are 180° out of phase
5.
For capacitor:

The charge decays exponentially with time. When the capacitor discharge.
For Inductor:

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