Question

A 10 V moving iron ammeter has a full scale deflection of 40mA on DC. Circuit . It reads 0.8% low on 50 Hz AC, Calculate the inductance

Answer 1

Answer:

Self inductance of the coil is 110 mH

Explanation:

When it is connected across 10 V DC source then it behave like a resistor

So its resistance is given as

$R = \frac{V}{i}$

$R = \frac{10}{40 \times 10^{-3}}$

$R = 250 ohm$

Now when it is connected across AC source then net impedence is given as

$z = \frac{10}{40(1-0.008)\times 10^{-3}}$

$z = 252 ohm$

now we have

$z^2 = x_L^2 + R^2$

$252^2 = x_L^2 + 250^2$

$x_L = 31.7$

$2\pi(50)L = 31.7$

$L = 110 mH$

#Learn

Topic : Impedence of LR coil

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