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Current senstivity of moving coil galvanometer is `5 "div"//mA` and its voltage senstivity (angular deflection per unit voltage applied) is `20 "div"//V`. The resistance of the galvanometer isA. `250 Omega`B. `25 Omega`C. `40 Omega`D. `500 Omega` |
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Answer» Correct Answer - A Current sensitivity of a moving coil galvanometer is the deflection `(theta)` per unit current (I) flowing through it, i.e. `I_(S) = theta/I = (NAB)/k` ……….(i) Where, N= numberof turns in the coil, A=Area of each turn of coil, B = magnetic field k= restoring torque per unit twist of the fibre stirp. Similarly, voltage sensitivity is the deflection per unit voltage. i.e., `V_(s) = theta/V = (NAB)/(k)(I/V) = (NAB)/(kR_(G)`............(ii) where, `R_(G)` is the resistance of the galvanometer. From eq(s) (i) and (ii), we get `R_(a) = I_(S)/V_(S)` ...............(ii) Here, `I_(s) = 5"div"/mA = 5 xx 10^(-3)"div"/A` and `V_(B)`=20 "div"/V Substituting the given values in Eq. (iii), we get `R_(a) = (5 xx 10(3))/20 = 250` `therefore` The resistance of the galvanometer is `250 Omega` |
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