JEE Main 2019 (Online) 12th April Evening Slot | Current Electricity Question 211 | Physics

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 214 English

JEE Main 2019 (Online) 12th April Morning Slot Physics - Current Electricity Question 214 English

16 $$\Omega $$

1 $$\Omega $$

8 $$\Omega $$

6 $$\Omega $$

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

A galvanometer of resistance 100 $$\Omega $$ has 50 divisions on its scale and has sensitivitv of 20 $$\mu $$A/division. It is to be converted to a voltmeter with three ranges of 0-2V, 0-10 V and 0-20 V. The appropriate circuit to do so is

JEE Main 2019 (Online) 12th April Morning Slot Physics - Current Electricity Question 213 English Option 1

JEE Main 2019 (Online) 12th April Morning Slot Physics - Current Electricity Question 213 English Option 2

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

To verify Ohm's law, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained : JEE Main 2019 (Online) 12th April Morning Slot Physics - Current Electricity Question 212 English

JEE Main 2019 (Online) 12th April Morning Slot Physics - Current Electricity Question 212 English

If V₀ is almost zero, identify the correct statement :

The value of the resistance R is 1.5 $$\Omega $$

The emf of the battery is l.5 V and its internal resistance is 1.5 $$\Omega $$

The emf of the battery is l.5 V and the value of R is 1.5 $$\Omega $$

The potential difference across the battery is 1.5 V when it sends a current of 1000 mA

PREVIOUS NEXT

JEE Main Subjects