Question

A battery of emf 1.4 v and internal resisyance 2 ohm is connected to a resistor of 100 ohm through an ammeter is 4/3 ohm.A voltmeter has also been connected to find the potential difference across the resistor. If the ammeter reads 0.02 a, then what will be the resistance of the voltmeter?

Answer 1

Answer:

Explanation:

Hey,

Here is the solution of your query..

=>Suppose, voltmeter has the resistance R ohm.

So, for voltmetre, the equivalent resistance R ohm and 100 ohm in parallel is

R*100/R + 100 = 100R/ R + 100

=> Here, A battery is connected to a resistor through an ammeter

and the resistance of ammeter = 4/3 ohm is given.

so, 100R/R + 100 + 4/3 + 2Ω

=>  The ammeter reads 0.02 A, therefore

I = V/R

$0.02 =\frac{1.4}{\frac{100R}{100+R}+\frac{4}{3}+2}$

R = 200 ohm

Thus, the resistance of the voltmeter will be 200 ohm.