A standard cell of emf `e_(0)=1.11` V is balanced against 72 cm length of a pootentiometer. The same potentiometer is used to measure the potential difference across the stndard resistance `R=120 Omega`. When the ammeter shows a current of 7.8 mA, a balanced lengh of 60 xm is obtained on the potentiometer.

(i) Determine the current following through the resistor.
(ii) Estimate the error in measurement of the ammeter.

Explanation

Here, `l_(0)=72cm, l=60cm, R=120Omega" and "epsi_(0)=1.11V`
(i) By using equation `epsi=xl_(0)…………………(i)
`V=lR=xl…………………(ii)
From equation (i) and (ii)
`1=epsi_(0)/R(l/l_(0))" "thereforeI=(1.11)/(120)(60/72)=7.7mA`
(ii) Since the measured reading 7.8 mA `(gt7.7 mA)` therefore, the instrument has a positive error.
`DeltaI=7.8-7.7=0.1 mA, (DeltaI)/I=0.1/7.7xx100=1.3%`