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I have the following problem:

A one-phase voltage of v(t) = 150⋅sin(wt)
is connected to a load with impedance Z=2+4j Ohm.
What is the current through the load, on phasor form? (use the supply voltage as reference)

My problem:
I think the magnitude is 23.7 (taking 150/sqrt(2) and dividing by 4.47 since r=sqrt(2^2 + 4^2). But how do i calculate the angle for the phasor form?

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  • \$\begingroup\$ It is usually good practice to write “rms” out beside rms quantities too. Avoids assumption errors. \$\endgroup\$ – relayman357 Jan 22 at 14:40
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Yes, the magnitude of the current is 23.72 amps.

To calculate the phase angle, the easiest way is to convert the rectangular impedance to a magnitude (already done) and an angle. The angle is \$\arctan(4/2)\$ = 63.435 °.

Given that it is a positive value, when you divide voltage (at 0 °) by an impedance having an angle of +63.435 ° the net result for current is a negative angle of 63.435 °

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