1
MHT CET 2021 23th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If we increase the frequency of an a.c. supply, then inductive reactance

A
increases directly with the square of frequency
B
increases as it directly proportional to frequency
C
decreases inversely with the square of frequency
D
decreases as it is inversely proportional to the frequency
2
MHT CET 2021 22th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

A capacitor of capacity '$$C$$' is charged to a potential '$$V$$'. It is connected in parallel to an inductor of inductance '$$\mathrm{L}$$'. The maximum current that will flow in the circuit is

A
$$V \sqrt{\frac{L}{C}}$$
B
$$\mathrm{V} \sqrt{\mathrm{LC}}$$
C
$$V \sqrt{\frac{C}{L}}$$
D
$$\frac{\mathrm{VC}^2}{\mathrm{~L}}$$
3
MHT CET 2021 22th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

A step down transformer is used to reduce the main supply from '$$V_1$$' volt to '$$V_2$$' volt. The primary coil draws a current '$$\mathrm{I}_1$$' $$\mathrm{A}$$ and the secondary coil draws '$$\mathrm{I}_2$$' A. $$(\mathrm{I}_1<\mathrm{I}_2)$$. The ratio of input power to output power is

A
$$\frac{V_1 V_2}{I_1 I_2}$$
B
$$\frac{V_1 I_1}{V_2 I_2}$$
C
$$\frac{\mathrm{I}_1 \mathrm{I}_2}{\mathrm{~V}_1 \mathrm{~V}_2}$$
D
$$\frac{\mathrm{V}_1 \mathrm{I}_2}{\mathrm{~V}_2 \mathrm{I}_1}$$
4
MHT CET 2021 22th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

For the circuit shown below, instantaneous current through inductor '$$\mathrm{L}$$' and capacitor '$$\mathrm{C}$$' is respectively.

MHT CET 2021 22th September Evening Shift Physics - Alternating Current Question 15 English

A
$$\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \cos \omega \mathrm{t} ; \mathrm{e}_0 \omega \mathrm{c} \cos \omega \mathrm{t}$$
B
$$\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \sin \omega t ; \frac{\mathrm{e}_0}{\omega \mathrm{C}} \cos \omega \mathrm{t}$$
C
$$\frac{\mathrm{e}_0 \mathrm{C}}{\mathrm{L}} \cos \omega \mathrm{t}; \frac{\mathrm{e}_0 \mathrm{~L}}{\mathrm{C}} \sin \omega \mathrm{t}$$
D
$$\frac{-\mathrm{e}_0 \mathrm{C}}{\mathrm{L}} \sin \omega \mathrm{t} ; \frac{\mathrm{e}_0 \mathrm{~L}}{\mathrm{C}} \cos \omega \mathrm{t}$$
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