A `50 Omega` resistance is connected to a battery of 5 V. A galvanometer of resistance `100 Omega` is to be used as an ammeter to measure current through the resistance, for this a resistancer `r_(s)` is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of the current without the ammeter in the circuit?

Options

A. `r_(s) = 0.5 Omega` in parallel with the galvanometer
B. `r_(s) = 0.5 Omega` in series with the galvanometer
C. `r_(s) = 1 Omega` in series with galvanometer
D. `r_(s) = 1 Omega` in parallel with galvanometer

Answer

a

Explanation

When there is no ammeter in the circuit, then current can be written as follows:
`I = V/R = 5/50 = 0.1 A`.
When ammeter is added in the circuit then current of the circuit decreases and it is given that current should not change by more than 1%. Hence current in the circuit when ammeter is included can be written as
`I. = 0.099 A`
Small variation in current is possible only when effective resistance of the galvanometer is minimum, So we should connect resistance S in parallel to the galvanometer.
Effective resistance of the circuit can be written as follows:
`R_("eff") = 50 + (100S)/(100 + S) = V/(I.)`
`rArr (100S)/(100 + S) = 5/(0.099) - 50 = 50.50 - 50`
`rArr (100S)/(100 + S) = 5/(0.099) - 50 = 50.50 - 50 = 0.5`
`rArr 100S - 0.5 (100 + S)`
` rArr 100S = 50 + 0.5S`
`rArr S = 0.5 Omega`
Hence `0.5 Omega` resistance should be connected in parallel to the coil of galvanometer. Hence option (a) is correct.