The emf `E` and the internal resistance r of the battery shown in figure are `4.3 V` and `1.0 Omega` respectively. The external resistance `R` is `50 Omega`. The resistances of the ammeter and voltmeter are `2.0 Omega` and `200 Omega` respectively.
(a) Find the readings 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`