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Question
Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω, N1 = 30,
A1 = 3.6 × 10–3 m2, B1 = 0.25 T
R2 = 14 Ω, N2 = 42,
A2 = 1.8 × 10–3 m2, B2 = 0.50 T
(The spring constants are identical for the two meters).
Determine the ratio of
- current sensitivity and
- voltage sensitivity of M2 and M1.
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Solution
For moving coil meter M1:
Resistance, R1 = 10 Ω
Number of turns, N1 = 30
Area of cross-section, A1 = 3.6 × 10–3 m2
Magnetic field strength, B1 = 0.25 T
Spring constant K1 = K
For moving coil meter M2:
Resistance, R2 = 14 Ω
Number of turns, N2 = 42
Area of cross-section, A2 = 1.8 × 10–3 m2
Magnetic field strength, B2 = 0.50 T
Spring constant, K2 = K
(a) Current sensitivity of M1 is given as:
`"I"_("s"_1) = ("N"_1"B"_1"A"_1)/"K"_1`
And, current sensitivity of M2 is given as:
`"I"_("s"_2) = ("N"_2"B"_2"A"_2)/"K"_2`
∴ Ratio `"I"_("s"_2)/"I"_("s"_1) = ("N"_2"B"_2"A"_2"K"_1)/("N"_1"B"_1"A"_1"K"_2)`
= `(42 xx 0.5 xx 1.8 xx 10^-3 xx "K")/(30 xx 0.25 xx 3.6 xx 10^-3 xx "K")`
= 1.4
Hence, the ratio of current sensitivity of M2 to M1 is 1.4.
(b) Voltage sensitivity for M2 is given as:
`"V"_("s"_2) = ("N"_2"B"_2"A"_2)/("K"_2"R"_2)`
And, voltage sensitivity for M1 is given as:
`"V"_("s"_1) = ("N"_1"B"_1"A"_1)/("K"_1"R"_1)`
∴ Ratio `"V"_("s"_2)/"V"_("s"_1) = ("N"_2"B"_2"A"_2"K"_1"R"_1)/("N"_1"B"_1"A"_1"K"_2"R"_2)`
= `(42 xx 0.5 xx 1.8 xx 10^-3 xx "K" xx 10)/(30 xx 0.25 xx 3.6 xx 10^-3 xx "K" xx 14)`
= 1
Hence, the ratio of voltage sensitivity of M2 to M1 is 1.
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