JEE Main 2021 (Online) 25th February Morning Shift | Alternating Current Question 119 | Physics

The current (i) at time t = 0 and t = $$\infty $$ respectively for the given circuit is :

JEE Main 2021 (Online) 25th February Morning Shift Physics - Alternating Current Question 119 English

JEE Main 2021 (Online) 25th February Morning Shift Physics - Alternating Current Question 119 English

$${{18E} \over {55}},{{5E} \over {18}}$$

$${{5E} \over {18}},{{18E} \over {55}}$$

$${{5E} \over {18}},{{10E} \over {33}}$$

$${{10E} \over {33}},{{5E} \over {18}}$$

JEE Main 2020 (Online) 6th September Morning Slot

MCQ (Single Correct Answer)

-1

Out of Syllabus

An AC circuit has R= 100 $$\Omega $$, C = 2 $$\mu $$F and L = 80 mH, connected in series. The quality factor of the circuit is :

0.5

400

JEE Main 2020 (Online) 4th September Evening Slot

MCQ (Single Correct Answer)

-1

A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then the time at which the energy stored in the inductor reaches $$\left( {{1 \over n}} \right)$$ times of its maximum value, is :

$${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n + 1}}} \right)$$

$${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)$$

$${L \over R}\ln \left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)$$

$${L \over R}\ln \left( {{{\sqrt n - 1} \over {\sqrt n }}} \right)$$

JEE Main 2020 (Online) 3rd September Morning Slot

MCQ (Single Correct Answer)

-1

A 750 Hz, 20 V (rms) source is connected to a resistance of 100 $$\Omega $$, an inductance of 0.1803 H and a capacitance of 10 $$\mu $$F all in series. The time in which the resistance (heat capacity 2 J/^oC) will get heated by 10^oC. (assume no loss of heat to the surroudnings) is close to :

348 s

418 s

245 s

365 s

PREVIOUS NEXT

Questions Asked from Alternating Current (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions