Current Electricity and Electrical InstrumentMedium
Question
A galvanometer has a resistance of 20Ω and reads full - scale when 0.2 V is applied across it. To convert it into a 10A ammeter, the Galvanometer coil should have a
Options
A.0.01Ω resistor connected across it
B.0.02Ω resistor connected across it
C.200Ω resistor connected across it
D.2000Ω resistor connected across it
Solution
V = IR ⇒ 0.2 = I(20)
Ig = 0.01A (through the galvanometer)
Ig G = (i - ig) S ⇒ (0.01) (20) = (10 - 0.01)S
⇒ S = 0.020Ω
Ig = 0.01A (through the galvanometer)
Ig G = (i - ig) S ⇒ (0.01) (20) = (10 - 0.01)S
⇒ S = 0.020Ω
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