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Two moving coil metres M(1) and M(2) hav...

Two moving coil metres `M_(1) and M_(2)` have the following particular `R_(1) =10Omega, N_(1) =30, A_(1) = 3.6 xx 10^(-3) m^(2), B_(1) =0.25 T,`
`R_(2) = 14Omega , N_(2) =42, A_(2) = 1.8 xx 10^(-3) m^(2), B_(2) =0.50 T`
The spring constants are identical for the two metres. What is the ratio of current sensitivity and voltage sensitivity of `M_(2)" to" M_(1)?`

Text Solution

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Current sensitivity of a moving coil metre of number of turns N, area A, magnetic field B and spring constant k is given by
Current sensitivity ` = (NBA)/k`
(a) Current sensitivity of `M_(1) = (N_(1)B_(1)A_(1))/k`
Current sensitivity of `M_(2) = (N_(2)B_(2)A_(2))/k`
` :. ("Current sensitivity of " M_(2))/("Current sensitivity of " M_(1)) = (N_(2)B_(2)A_(2))/(N_(1)B_(1)A_(1))`
Here,
` N_(1) = 30, A_(1) = 3.6 xx 10^(-3) m^(2) , B_(1) = 0.25 T`
` N_(2) = 42 , A_(2) = 1.8 xx 10^(-3)m^(2), B_(2) = 0.50 T`
` rArr ("Current sensitivity of " M_(2))/("Current sensitivity of " M_(1)) = (42 xx 0.5 xx 1.8 xx 10^(-3))/(30 xx 0.25 xx 3.6 xx 10^(-3))`
` = 1.4`
(b) Also, voltage sensitivity if the resistance of the metre is R is given by
Voltage sensitivity ` = (NBA)/(kR) = "Current sensitivity"/R`
` :. ("Voltage sensitivity of "M_(2))/("Voltage sensitivity of " M_(1)) = ("Current sensitivity of " M_(2))/("Current sensitivity of " M_(1)) xx R_(1)/R_(2) `
Here, `R_(1) = 10 Omega " and " R_(2) = 14 Omega`
` :. ("Voltage sensitivity of " M_(2))/("Voltage sensitivity of " M_(1)) = 1.4 xx 10/14 = 1`
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