Q. 104.1( 21 Votes )

# Two moving coil meters, M_{1} and M_{2} have the following particulars:

R_{1} = 10Ω, N_{1} = 30,

A_{1} = 3.6 × 10^{–3} m^{2}, B_{1} = 0.25 T

R_{2} = 14Ω, N_{2} = 42,

A_{2} = 1.8 × 10^{–3} m^{2}, B_{2} = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M_{2} and M_{1}.

Answer :

Given:

For moving coil meter M_{1}

Resistance of wire, R_{1} = 10Ω

Number of turns, N_{1} = 30

Area of cross-section, A_{1} = 3.6 × 10^{-3} m^{2}

Magnetic field strength, B_{1} = 0.25 T

For moving coil meter M_{2}

Resistance of wire, R_{2} = 14Ω

Number of turns, N_{2} = 42

Area of cross-section, A_{2} = 1.8 × 10^{-3} m^{2}

Magnetic field strength, B_{2} = 0.50 T

Spring constant, K_{1} = K_{2} = K

a) Current sensitivity is given by,

For M_{1},

…(1)

By putting the values in equation 1, we have

⇒

For M_{2}

…(2)

By putting the values in equation 2, we have

⇒

Now finding the ration of I_{1} and I_{2}

⇒

Hence, the ratio of current sensitivities is 1.4.

b) Voltage sensitivity is given by,

For M_{1},

⇒ …(3)

For M_{2}

⇒ …(4)

By dividing equation 3 by 4, we get

⇒

Hence, the ratio of voltage sensitivity of M_{1} and M_{2} is 1.

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