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) 0.1A, 4.0 V,
(b) `(1083.6xx200)/(10652xx252)=0.08A, 4.3-(1083.6)/(10652)=4.2V`