Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω, N1 = 30,
A1 = 3.6 x 10-3 m2, B1 = 0.25 T
R2 = 14 Ω, N2 = 42,
A2 = 1.8 x 10-3 m2, B2 = 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.
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:
And, current sensitivity of M2 is given as:
so, Ratio
Hence, the ratio of current sensitivity of M2 to M1 is 1.4.
(b) Voltage sensitivity for M2 is given as:
And, voltage sensitivity for M1 is given as:
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 10: Two moving coil meters, M1 and M2 have the following particulars: R1 = 10 Ω, N1 = 30, A1....
Comments
Not clear can't understand
Not clear