The emf `(epsilon)`and the internal resistance r of the battery shown in figure are 4.3V and `1.0 Ohm` respectively.The external resistance R is `50 ohm`.The resistances of the ammeter and voltmeter are `2.0 ohm and `200ohm respectively.(a)Find the reading of the two meters.(b) The switch is thrown to the other side, What will be the readings of the two meters now?

Answer

A, B, D

Explanation

a.
`R_("net")=1.0+20.+(50xx200)/(50+200)`
`=43 Omega`
`:. i=4.3/4.3=0.1A`
= Readings of ammeter
Readings of voltmeter
= (i) net resistance of `50 Omega` and `200 Omega`
`=(0.1)((50xx200)/(50+200))`
`=4V`
b.

`R_("net")=1.0+(52xx200)/(52+200)=42.27Omega`
`:. i=4.3/42.27=0.1A`
Now `i_1/i_2=200/52`
`:.i_1=(200/252)(0.1)=0.08A`
=Reading of ammeter
`:.` Reading to voltmeter
=Potential difference across `50 Omega` and `2.0 Omega`
`=0.08xx52=4.2V`