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A galvanometer with a coil of resistance 120 ohm shows full scale full scale deflection for a current of 2.5 mA. How will you convert the galvonometer into an ammeter of range 0 to 7.5 A ? Determine the net resistance of the ammeter. When an ammeter is put in a circuit, does it read slightly less or more than the actual current in the original circuit ? Justify your answer. |
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Answer» Solution :Here `R_G = 120 Omega, I_g = 2.5 mA = 2.5 xx 10^(-3) A and I = 7.5 A` `:.` Shunt resistance NEEDED to converted galvanometer into AMMETER `r_s = (I_g)/(I - I_g) cdot R_G = (2.5 xx 10^(-3) xx 120)/(7.5 - 2.5 xx 10^(-3)) = 0.04 Omega` `:.` Net resistance of the ammeter `R = (G r_S)/(G + r_S) = (120 xx 0.04)/(120 + 0.04) = 0.04 Omega` when an ammeter is put in an electric CIRCUIT, it reads slightly less becase net resistance of the circuit slightly rises due to resistance of ammeter being included in series. CONSEQUENTLY, the current slightly decreases. |
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