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A moving-coil galvanometer of resistance 200 ohms gives full scale deflection of 100 divisons for a current of 50 milliamperes. How will you convert it into an ammeter to read 2 amperes for 20 divisons? |
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Answer» Solution :Data: `G=200 Omega, I_(g) =50 mA = 50 xx 10^(-3)A` The total NUMBER of scale divisions is 100. The ammeter has to read 20 divisons for a CURRENT of 2A. HENCE, for 100 divisions, the current MUST be I=10A. To convert the galvanometer into an ammeter, a low resistance (shunt) must be connected in parallel with the galvanometer coil. The required resistance, `S=(I_(g)/(I-I_(g)))G` `=(50 xx 10^(-3))/(10-50 xx 10^(-3)) xx 200` `=(50 xx 10^(-3))/((10000 -50) xx 10^(-3)) xx 200` `=(200)/(200-1)= 200/199 = 1.005 Omega` |
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