Two ammeters, 1 and 2 have different internal resistances: `r_(1)` (known) and `r_(2)` (known). Each ammeter has scale such that the angular deviation of the needle for zero is proportional to the current passing through it. Initially, the ammeters are connected in series to a source of constant voltage. The deviations of the needles of the ammeter are `theta_(1)` and `theta_(2)`, respecitively. The ammeters are then connected in parallel ot the same voltage source. This time, the deviations of the needles are `theta'_(1)` and `theta'_(2)` respectively. Find `r_(2)` in terms of `r_(1), theta_(1), theta_(2), theta'_(1)` and `theta'_(2)`.

Options

A. `r_(1) (theta_(1) theta'_(1))/(theta_(2) theta'_(2))`
B. `r_(1) (theta_(2) theta'_(1))/(theta_(1) theta'_(2))`
C. `r_(1) (theta_(2) theta'_(2))/(theta_(1) theta'_(1))`
D. `r_(1) (theta_(2) theta'_(2))/(theta_(1) theta'_(2))`

Answer

B

Explanation

(b) In first case
`i= k_(1), theta_(1), i= k_(2) theta_(2) k_(1), theta_(1) = k_(2), theta_(2)`
In second case,
`V_(1) = V_(2) implies k_(1) theta'_(1) r_(1) = k_(2) theta'_(2) r_(2)`
from (1) and (2), we get `r_(2) = r_(1) (theta_(2) theta'_(1))/(theta_(1) theta'_(2))`