Question

An ammetre of resistance 100 ohm can measure a maximum current of 5 mA what will do to measure the maximum current of 5 A with it ?​

Answer 1

Answer:

Given: a galvanometer of resistance 100Ω which can measure a maximum current of 1mA is converted into an ohmmeter by connecting a battery of emf 1V and a fixed resistance of 900Ω in series. When an external resistance is measured the current reading is 0.1mA.

To find the value of resistance

Solution:

According to the given criteria,

I

g

=1mA,V=1V,G=100Ω,R=900Ω,G+R=1000Ω

Hence,

I=

R

ex

V

=

1000

1

=10

−3

A

And, I

′

=

R

ex

V

⟹R

ex

=

I

′

V

⟹

10

−4

1

⟹R

ex

=10

4

Ω

Therefore, G+R+R

′

=10,000Ω

Therefore, R

′

=10000−1000=9000Ω

is the required value of the resistance.

Answer 2

Answer:

resistance to drop 10 volts at 5 ma.

R=E/I, R=10/.005=2000 ohms,

then subtract the meters internal resistance,

2000–100=1900.

So 1900 ohms is the required resistance in series with the meter. This may require two or more resistors.