- #1

Tusike

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## Homework Statement

Hi!

We have a parallel circuit on 230V, 60Hz voltage. One of the parallel's has a 0.0768H coil, the other has a 50 Ohm resistance on it. The question is: What is the effective I in the main current?

## Homework Equations

Z = Lw

## The Attempt at a Solution

OK, so I have two ways of solving this, and they lead to different results, I don't know which isn't correct and why.

#1). The impedance of the resistance is the same, 50 Ohms, while the impedance of the coil is L*w = 0.0768H * 120*pi = 28.868 Ohms. And so for the overall impedance: 1/Z = 1/Z(R) + 1/Z(L) = 1/50 + 1/28.868 = 0.05464 and so Z = 18.3 Ohms. For the effective I in the main current; I(eff) = U/Z = 230V/18.3Ohms = 12.567A.

#2) Since the coil's current is 90 degrees behind the resistance's, it is true that the main current i^2 = i(ZL)^2 + i(ZR)^2, where i(ZL) is the current in the coil and i(ZR) is the current in the resistance at a given point in time. Divide this equation by sqr(2)*sqr(2) to get the same equation for the effective currents: I^2 = I(ZL)^2 + I(ZR)^2.

We have calculated before that Z(L) = 28.868, so I(ZL) = 230/28.868=7.9673A. Z(R) = R = 50, so I(ZR) = 230 / 50 = 4.6A. And so the effective current in the main current is sqr(7.9672^2 + 4.6^2) = 9.2A.

Can anybody please tell me what I did wrong and why is it wrong? Thanks!

-Tusike