An ammeter and a voltmeter are initially connected in series to a battery of zero internal resistance. When switch `S_(1)` is closed the reading of the voltmeter becomes half of the initial, whereas the reading of the ammeter becomes double. If now switch `S_(2)` is also closed, then reading of ammeter becomes :

Options

A. 3/2 times the initial value
B. 3/2 times the value after closing `S_(1)`
C. 3/4 times the value after closing `S_(1)`
D. 3/4 times the initial value

Answer

B

Explanation

Initially :-
`V_(v) + V_(A) = 6`….(1)
`V_(v)` & `V_(A)` being the potential across voltmeter & ammeter respectively
after closing `S_(1)`
`(V_(v))/(2) +2 V_(A)=6`...(2)
Solving (1) & (2)
`V_(v) = 4,V_(A)=2`
after closing `S_(2)` :-
`V_(v) =0`
`V_(A) = 6`
So that value after closing `S_(2)` is `3//2` times the value after closing `S_(1)`