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Two moving-coil galvanometer, P and Q, are alike in all respects except that P's coil has 10 turns of resistance 2Omega and Q's coil has 100 turns of resistance 30 Omega. Compare their (i) current sensitivities (ii) Voltage sensitivities. |
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Answer» SOLUTION :DATA: `N_(p) = 10, G_(p)= 2Omega, N_(Q) = 100, G_(Q) = 30 Omega` (i) The current sensitivity, `S_(1) = (NAB)/C` `therefore S_(LP)/S_(LQ) = N_(P)/N_(Q) = 10/100=1/10` (ii) The voltage sensitivity, `S_(v) = S_(l)/G` `therefore S_(V.P)/S_(I.Q).G_(Q)/G_(P)=1/(10.30/2)=1.5` |
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