Question

$\bf \underline{Question} \: \: :$

➝ The moving coil meters, $\sf M_1$ and $\sf M_2$ have the following particulars
$\sf R_1 = 10Ω, N_1 = 30$
$\sf A_1 = 3.6 × 10^{-3} m^{2}, B_1 = 0.25T$
$\sf R_2 = 40 Ω, N_2 = 42$
$\sf A_2 = 1.8 × 10^{-3} m², B_2 = 0.50T$
(The spring constants are idenical for the two meters)

Determine the ratio of

(a) Current sensitivity and

(b) Voltage sensitivity of $\sf M_1$ and $\sf M_2$

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Answer 1

Given :-

N = 30 turns

R = 10Ω

A = 3.6 × 10-³ m²

B = 0.25 T

N' = 42 turns

R' = 40Ω

A' = 1.8 × 10-³ m²

B' = 0.5 T

It is also given that,

K = K' = K

As we know that,

Current Sensitive = I = NBA/K

To find the ratio we can Simply divide the values after putting them in formula,

I/I' = NBA/K × K'/N'B'A'

I/I' = 30 × 0.25 × 3.6 × 10-³/K × K/42 × 0.5 × 1.8 × 10-³

I/I' = 30 × 0.25 × 3.6 × 10-³/42 × 0.5 × 1.8 × 10-³

I/I' = 30/42

I/I' = 0.71

Again,

For Voltage Sensitivity,

V = NBA/KR

V/V' = 30 × 0.25 × 3.6 × 10-³/10K × 40K/42 × 0.5 × 1.8 × 10-³

V/V' = 30 × 40/42 × 10

V/V' = 2.85

Hence,

The ratio of Current sensitive = I/I' = 0.71

The ratio of Voltage Sensitivity = V/V' = 2.85

Answer 2

Answer:

Explanation:

The answer is 1.4 and 1