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InterviewSolution
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The coil of a moving coil galvanometer has 5 turns and each turn has an effective area of 2 xx 10^(-2) m^(2) . It is suspended in a magnetic field whose strength is 4 xx 10^(-2) Wb m^(-2) . If the torsional constant K of the suspension fibre is4xx 10^(9) N m deg^(-1) . (a) Find its current sensitivity in degree per micro - ampere (b) Calculate the voltage sensitivity of the galvanometer for it to havefull scale deflection of 50 divisions for 25 mV. (c ) Compute the resistance of the galvanometer . |
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Answer» Solution :The number of turns of coil is 5 turns The area of each coil is ` 2xx 10^(-2) m^(2)` Strength of the magnetic FIELD is 4 `xx 10^(-2) ` Wh `m^(-2)` Torsional constant is 4 `xx 10^(-9) ` N m `deg^(-10` `I_(x) = ("N A B")/( K)= (5 xx 2 xx 10^(-2) xx 4 xx 10^(-2))/( 4 xx 10^(-9)) = 10^(@)` divisions per ampere `I mu`A = 1 microampere = `10^(-6)` ampere THEREFORE, ` I_(x)= 10^(6) ("div")/(A) = 1 ("div")/(10^(-6) A) = 1 ("div")/(mu A)` `I_(x) ` = 1div `(mu A)^(-1)` (b) Voltage sensitivity `V_(x)= (theta)/(V) = (50"div")/(25 m V) = 2xx 10^(3) "div" V^(-1)` (c) The resistance of the galvanometer is `R_(g) = (I_(s))/(V_(s)) = (10^(6) ("div")/(A))/( 2 xx 10^(3) ("div")/(A)) = 0.5 xx 10^(3) (V)/(A) = 0.5 k Omega` |
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