JEE Main 2024 (Online) 1st February Evening Shift | Current Electricity Question 26 | Physics

In an ammeter, $5 \%$ of the main current passes through the galvanometer. If resistance of the galvanometer is $\mathrm{G}$, the resistance of ammeter will be :

$199 \mathrm{~G}$

$200 \mathrm{~G}$

$\frac{G}{20}$

$\frac{\mathrm{G}}{199}$

JEE Main 2024 (Online) 1st February Morning Shift

MCQ (Single Correct Answer)

-1

The reading in the ideal voltmeter $(\mathrm{V})$ shown in the given circuit diagram is :

JEE Main 2024 (Online) 1st February Morning Shift Physics - Current Electricity Question 47 English

JEE Main 2024 (Online) 1st February Morning Shift Physics - Current Electricity Question 47 English

$3 \mathrm{~V}$

$10 \mathrm{~V}$

$5 \mathrm{~V}$

$0 \mathrm{~V}$

JEE Main 2024 (Online) 1st February Morning Shift

MCQ (Single Correct Answer)

-1

A galvanometer has a resistance of $50 ~\Omega$ and it allows maximum current of $5 \mathrm{~mA}$. It can be converted into voltmeter to measure upto $100 \mathrm{~V}$ by connecting in series a resistor of resistance :

$19500 \Omega$

$5975 \Omega$

$20050 \Omega$

$19950 \Omega$

JEE Main 2024 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

-1

By what percentage will the illumination of the lamp decrease if the current drops by 20%?

26%

36%

46%

56%