An ammeter and a voltmeter of resistance R are connected in series to an electric cell of negligible internal resistance . Their readings are A and V respectively . If another resistance R is connected in prallel with voltmeter

Options

A. Both A and V will increase
B. Both A and V will decrease
C. A will decrease and V will increase
D. A increase and V will decrease

Answer

C

Explanation

Current , `i= E/(R+r)`
Rightarrow `0.5= (E )/(R+r)`
Rightarrow `0.9 = (E )/(5+r)`
From equation (1) and (2)
`0.5 (11+r) = 0.9(5+r)`
Rightarrow `(11+r)/(5+r)= (9)/(5)`
therefore `r= 2.5Omega`