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A 240/110V transformer has a no-load primary current of 0.2A. When a purely resistive load is connected to the secondary, the primary current is 1.5A. Ignoring all losses and leakage effects, find the secondary current and draw a phasor diagram for the transformer.

My question is where does the -7.6 come from in this solution? I understand the other angles but unsure where this comes into effect.

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My question is where does the -7.6 come from in this solution.

It's actually -7.6622° and this is closer to -7.7°.

To find the purely resistive part of the current when loaded, you use Pythagoras: -

$$I_R = \sqrt{1.5^2 - 0.2^2} = 1.48661\text{ amps}$$

To find the angle of the 1.5 amps taken by the primary circuit, use the inverse tan function: -

$$\theta = \tan^{-1}\left(\dfrac{-0.2}{1.48661}\right) = -7.6622°$$

Or you could use \$\sin^{-1}\left(\dfrac{-0.2}{1.5}\right)= -7.6622°\$

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