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A moving coil galvonometer of resistance R_G gives a full scale deflection for current I_g. Use the suitable circuit diagram convert it into an ammeter of range 0 to I (I > I_g). Deduce the expression for the shunt required for this conversion. Hence, write the expression for the resistance of the ammeter thus obtained. |
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Answer» Solution :Let a MOVING coil galvonometer of resistance `R_G`, gives FULL scale DEFLECTION for a current `I_g`. To convert it into an ammeter of range I, we join a shunt resistance of suitable value `r_s` so that out of total current I only a part `I_g` passes through the galva-nometer and rest of the current is passed through the shunt. It means that `I_g = (r_s)/(r_s + r_G) cdot I` `implies r_s = R_G cdot (I_g)/(I - I_g) = (R_G)/(((I)/(I_g) - 1)) = (R_G)/(N-1) "[ where " n = I/(I_g)]` As in an ammeter resistance of galvanometer `R_G` and shunt resistance `r_s` are joined in parallel , the net resistance of the ammeter `(r_A)` is given by `r_A = (R_G cdot r_s)/((R_G + r_s)) `, which is even LESS than the shunt resistance `r_s`. 
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