Question

An ammeter together with an unknown resistance in series is connected across two identical batteries each of emf 1.5 V. When the batteries are connected in series, the galvanometer records a current of 1A and when the batteries are in parallel, the current is 0.6A. Then the internal resistance of the battery is ______.

पर्याय

• `1/4 Omega`

• `1/5 Omega`

• `1/3 Omega`

• `1/2 Omega`

Answer 1

An ammeter together with an unknown resistance in series is connected across two identical batteries each of emf 1.5 V. When the batteries are connected in series, the galvanometer records a current of 1A and when the batteries are in parallel, the current is 0.6A. Then the internal resistance of the battery is `underline(1/3 Omega)`.

Explanation:

Internal resistances are in series when the batteries are connected in series. Internal resistances and emf both add up.

I = `"E"/("r + R")`

1 = `3/(2"r" + "R") => "R + 2r" = 3`   ...(1)

and in parallel combination, 

`0.6 = (2 xx 1.5)/("r"/2 + "R") => 5 = "2R" + "r"`  ...(ii)

From equation (i) and (ii),

r = `1/3  Omega`