Class 12 Physics
CBSE and UP Board
Moving Charges and Magnetism
Chapter-4 Exercise 4.28
Additional Exercise
A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?
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Resistance of the galvanometer coil, G = 15 Ω
Current for which the galvanometer shows full scale deflection,
Ig= 4mA = 4 x 10⁻³ A
Range of the ammeter is 0, which needs to be converted to 6 A.
Therefore ,Current,I = 6 A
A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter. The value of S is given as:
S =(Ig G)/(I -Ig) = (4 x 10⁻³ x 15)/(6 – 4 x 10⁻³)
=(6 x 10⁻² )/(6 – 0.004) = 0.06 /5.996
≈ 0.01 Ω = 10 mΩ
Hence, a shunt resistor is to be connected in parallel with the galvanometer.