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Determine the Ratio of (A) Current Sensitivity and (B) Voltage Sensitivity of M2 And M1. - Physics

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Two moving coil meters, M1 and M2 have the following particulars:

R1 = 10 Ω, N1 = 30,

A1 = 3.6 × 10–3 m2B1 = 0.25 T

R2 = 14 Ω, N2 = 42,

A2 = 1.8 × 10–3 m2B2 = 0.50 T

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

Determine the ratio of (a) current sensitivity and (b) 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_(s1)=(N_1B_1A_1)/K_1`

And, current sensitivity of M2 is given as:

`I_(s2)=(N_2B_2A_2)/K_2`

∴ Ratio `I_(s2)/I_(s1)=(N_2B_2A_2)/(N_1B_1A_1)  K_1/K_2`

`=(42xx0.5xx1.8xx10^-3xxK)/(Kxx30xx0.25xx3.6xx10^-3)=1.4`

Hence, the ratio of current sensitivity of M2 to M1 is 1.4.

(b) Voltage sensitivity for M2 is given as:

`V_(s2)=(N_2B_2A_2)/(K_2R_2)`

And, voltage sensitivity for M1 is given as:

`V_(s1)=(N_1B_1A_1)/(K_1R_1)`

∴ Ratio `v_(s2)/v_(s1)=(N_2B_2A_2)/(N_1B_1A_1)  (K_1R_1)/(K_2R_2)`

`=(42xx0.5xx1.8xx10^-3xx10xxk)/(Kxx14xx30xx0.25xx3.6xx10^-3)=1`

Hence, the ratio of voltage sensitivity of M2 to M1 is 1.

Concept: Moving Coil Galvanometer
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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 4 Moving Charges and Magnetism
Q 10 | Page 169

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