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Question
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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Solution
Resistance of the galvanometer coil, G = 15 Ω
Current for which the galvanometer shows full scale deflection, Ig = 4 mA = 4 × 10−3 A
Range of the ammeter is 0, which needs to be converted to 6 A.
∴ 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 = `("I"_"g""G")/("I" - "I"_"g")`
= `(4 xx 10^-3 xx 15)/(6 - 4 xx 10^-3)`
S = `(6 xx 10^-2)/(6 - 0.004)`
= `0.06/5.996`
≈ 0.01 Ω = 10 mΩ
Hence, a 10 mΩ shunt resistor is to be connected in parallel with the galvanometer.
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