Question

explain how a galvanometre with resistance 'G' and current at full scale deflection 'Ig' is converted into____
a) Voltmetre of range (0-V) (volt)
b)Ammetre of range (0-I) amphere​

Answer 1

Explanation:

.............................

Answer 2

Answer:

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

I

P

g=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 I

g

=0.3A and (I−I

g

) should pass through S

(I−I

g

)S=I

g

G

∴S=

(I−I

g

)

I

g

G

=

(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 I

a

=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=I

g

(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

s

) is connected in series with the galvanometer.

The equivalent meter resistance is

R

eq

=30+70=100ohm In this case, most of the voltage appears across the series resistor. The current through the voltmeter is 2mA

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