The emf `(epsilon)`and the internal resistance r of the battery shown in figure are 4.3V 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 reading of the two meters.(b) Tje switch is thrown to the other side, What will be the readings of the two meters now?

Explanation

(a) In circuit ABFGA,
` i_1 50 + 2i + i-4.3= 0 `
` rArr 50 i_1 + 3i = 4.3 ` ……….(i) `
` In circuit BEDCB, `
` rArr 50i_1-(i-i_1)xx 200 = 0 `
` rArr 50i_1 -200i +200i_1 = 0 `
` rArr 250 i_1 = 200i `
` rArr 50i_1 - 40i = 0 ` .........(ii)
` Solving equation (i) and (ii) we get `
`43i = 4.3 `
` rArr i = 0.1 `
` rArr 5i_1 = 4 xx i = 4 xx 0.1 `
` i_1 = 0.4/5 A. `
The current read by Ammeter, i = 0.1A
Voltameter reads a pot. diff.
` = i_1 xx 50 = 4V`
(b) In circuit ABEFA,
` 50i_1 + 2i_1 + 1i - 4.3= 0 `
` rArr 52i_1 + i - 4.3 = 0 `
` rArr 200(52i_1 + i - 4.3) = 0 `
` In circuit BCDEB, `
` (i-i_1)200 -2i-50i_1= 0 `
` rArr 200i -252i_1 = 0 `
` (10400+252) i_1 = 4.3 xx 2 xx 100 `
` rArr i_1(10652) = 4.3 xx 2 xx 100`
` rArr i=((4.3 xx 2 xx 100) /10652) = 0.08 `
` i = 4.3 - 52 xx 0.08 = 0.14. `
Reading of the ammeter = 0.08A
Reading of the voltameter
= (0.14 - 0.08)200
= 12 V .