Question

13. The resistance of a galvanometer is 30 2 and it
has 100 divisions on the scale. A current of 0.2 mA
produces a deflection of 1 division. What shunt will
be required for measuring a current of 5 A?​

Answer 1

ANSWER

Here, galvanometer resistance G=30Ω and full scale deflection current 

IPg=2mA

(a) To convert the galvanometer into an ammeter of range 0.3 ampere, a resistance of value S is connected in parallel with it such that the current through G should not be more than Ig=0.3A and (I−Ig) should pass through S

(I−Ig)S=IgG

∴S=(I−Ig)IgG=(0.3−2×10−3)2×10−3×30=0.2Ω

Hence, to convert the galvanometer into an ammeter of the desired range a shunt resistance (a small valued resistance) of 0.2Ω is connected parallel to the meter. This shunt resistance gives us a low resistance insturment with a deflection current Ia=0.3 ampere, while the current through the galvanometer is 2mA.

(b) To convert the galvanometer into a voltmeter of range 0.2 volt, a resistance R is connected in series with it such that

V=Ig(R+G)

0.2=2×10−3(30+R)

or R=100−30−70ohms

Thus, to convert the galvanometer into a voltmeter of the desired range, a high resistance (R)