JEE Main 2020 (Online) 7th January Evening Slot | Current Electricity Question 263 | Physics

In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value) of the building will be :

15 A

20 A

25 A

10 A

JEE Main 2020 (Online) 7th January Morning Slot

MCQ (Single Correct Answer)

-1

The current I₁ (in A) flowing through 1 $$\Omega $$ resistor in the following circuit is : JEE Main 2020 (Online) 7th January Morning Slot Physics - Current Electricity Question 265 English

JEE Main 2020 (Online) 7th January Morning Slot Physics - Current Electricity Question 265 English

0.4

0.25

0.2

0.5

JEE Main 2019 (Online) 12th April Evening Slot

MCQ (Single Correct Answer)

-1

A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I₀(I₀ > I_g) by connecting a shunt resistance R_A to it and (ii) into a voltmeter of range 0 to V (V = GI₀) by connecting a series resistance R_V to it. Then,

$${R_A}{R_V} = {G^2}$$ and $${{{R_A}} \over {{R_V}}} = {{{I_g}} \over {\left( {{I_0} - {I_g}} \right)}}$$

$${R_A}{R_V} = {G^2}\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)$$ and $${{{R_A}} \over {{R_V}}} = {\left( {{{{I_0} - {I_g}} \over {{I_g}}}} \right)^2}$$

$${R_A}{R_V} = {G^2}\left( {{{{I_0} - {I_g}} \over {{I_g}}}} \right)$$ and $${{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}$$

$${R_A}{R_V} = {G^2}$$ and $${{{R_A}} \over {{R_V}}} = {\left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)^2}$$

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is: JEE Main 2019 (Online) 12th April Morning Slot Physics - Current Electricity Question 269 English