MHT CET 2021 22th September Evening Shift | Alternating Current Question 44 | Physics

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

$$\frac{V_1 V_2}{I_1 I_2}$$

$$\frac{V_1 I_1}{V_2 I_2}$$

$$\frac{\mathrm{I}_1 \mathrm{I}_2}{\mathrm{~V}_1 \mathrm{~V}_2}$$

$$\frac{\mathrm{V}_1 \mathrm{I}_2}{\mathrm{~V}_2 \mathrm{I}_1}$$

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

-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 2 English

$$\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \cos \omega \mathrm{t} ; \mathrm{e}_0 \omega \mathrm{c} \cos \omega \mathrm{t}$$

$$\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \sin \omega t ; \frac{\mathrm{e}_0}{\omega \mathrm{C}} \cos \omega \mathrm{t}$$

$$\frac{\mathrm{e}_0 \mathrm{C}}{\mathrm{L}} \cos \omega \mathrm{t}; \frac{\mathrm{e}_0 \mathrm{~L}}{\mathrm{C}} \sin \omega \mathrm{t}$$

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

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

-0

A parallel plate capacitor having plates of radius 6 cm has capacitance 100 pF. It is connected to 230 V a.c. supply with angular frequency 300 rad/s. The r.m.s. value of current is

$$6.9\times10^{-6}$$ A

$$2.3\times10^{-5}$$ A

$$6.9\times10^{-5}$$ A

$$6.9\times10^{-7}$$ A

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

-0

A step down transformer has turns ratio $$20: 1$$. If $$8 \mathrm{~V}$$ is applied across $$0.4 \mathrm{~ohm}$$ secondary then the primary current will be

2 A

1 A

0.5 A

4 A

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