An ammeter and a voltmeter are joined in series to a cell. Their readigns are A and V respectively. If a resistance is now joined in parallel with the voltmeter, then

Options

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

Answer

B

Explanation

Ammeter reading (A) and voltmeter reading (V) depend on the circuit current and current through the voltmeter respectively.

Ist case: Circuit current, `I_(1)=(E)/(R_(A)+R_(V))`
`R_(A) and R_(V)` are the resistances of the ammeter and voltmeter respectively. the internal resistance (r) of the cell is neglected.
Current through voltmeter.
`I_(V_(1))=I_(1)=(E)/(R_(A)+R_(V))` . .(i)
2nd case: When a resistance R is connected in parallel with the voltmeter,
circuit current, `I_(2)=(E)/(R_(A)+(R R_(V))/(R+R_(V)))`
`becauseI_(2)gtI_(1)`, ammeter reading A will increase.
Now, current through voltmeter, `I_(V_(2))=(E)/(R+R_(V))*I_(2)`
or, `I_(V_(2))=(R)/(R_(V)+R)*(E)/(R_(A)+(R R_(V))/(R+R_(V)))`
`=(RE)/(R_(A)R_(V)+R R_(A)+R_(V)R)=(E)/((R_(A)R_(V))/(R)+R_(A)+R_(V))` . . (iii)
Comparing (i) and (ii), we get
`I_(V_(2)) lt I_(V_(1))`.
`therefore` voltmeter reading V will decrease.