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?

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 .