Question
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 k are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.
Solution
(a) Current sensitivity of a moving coil galvanometer is the amount of deflection per unit current.
Therefore,
Current sensitivity of first meter is given by
Is = ϕI= BANk
IM1 = ϕI = B1A1N1k =0.25 × 3.6 × 10-3 × 30k = 27 × 10-3k ...(i)
Current sensitivity of second meter,
IM2 = ϕI = B2A2N2k = 0.50 × 1.80 × 10-3 × 42k = 37.8 × 10-3k ...(ii)
Ratio of current sensitivity is therefore,
IM2IM1 =37.8 × 10-3k27 × 10-3k = 1.4
(b) Voltage sensitivity of a moving coil galvanometer is given as the amount of deflection per unit voltage.
Thus,
Voltage sensitivity of first meter is,
VM1= ϕV= ϕI. R = 27 × 10-3k × 10 = 2.7 ×10-3 k
Voltage sensitivity of second meter
VM2= ϕR.I = 37.8 × 10-3k × 10 = 2.7 × 10-3k
Hence, the ratio of voltage sensitivity, VM1VM2 = 1.
Therefore,
Current sensitivity of first meter is given by
Is = ϕI= BANk
IM1 = ϕI = B1A1N1k =0.25 × 3.6 × 10-3 × 30k = 27 × 10-3k ...(i)
Current sensitivity of second meter,
IM2 = ϕI = B2A2N2k = 0.50 × 1.80 × 10-3 × 42k = 37.8 × 10-3k ...(ii)
Ratio of current sensitivity is therefore,
IM2IM1 =37.8 × 10-3k27 × 10-3k = 1.4
(b) Voltage sensitivity of a moving coil galvanometer is given as the amount of deflection per unit voltage.
Thus,
Voltage sensitivity of first meter is,
VM1= ϕV= ϕI. R = 27 × 10-3k × 10 = 2.7 ×10-3 k
Voltage sensitivity of second meter
VM2= ϕR.I = 37.8 × 10-3k × 10 = 2.7 × 10-3k
Hence, the ratio of voltage sensitivity, VM1VM2 = 1.
