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A series RL is driven by a 120V, 60Hz source. The values of the resistane and inductance are R=20 ohm and L=160mH. Determine the current.

a. 0.86 b. 0.34 c. 1.89 d.4.32 e.0.35

I started off doing:

I = V/wL

I= 120 / (60)(160 x 10^-3)

But that's not right. What equation do I use then?

Hello Sona,

There are a couple problems with your attempt. You didn't include the resistor in your equation. The series RL has 120V across it, across the pair of components. The net impedance of the RL series combination, Xnet is the "sum" of the resistor's impedance and the inductor's impedance.

So the current will be

I = V / (Xnet)

As you know, the impedance of the inductor is w*L . But w is the angular frequency, you used the frequency, 60 Hz. You need to use 2*pi*f as the value of w.

So XL = 2*pi*f*L

I said above that the impedance of the series combination of the R and L is the "sum" of the resistor's impedance and the inductor's impedance. But here is where it gets complicated. Summing them is like using vector algebra to find the resultant vector of 2 vectors that are at an angle.

The inductor's impedance is at a 90 degree angle to the resistance. So using Pythagoras, the magnitude of the net impedance is

Xnet = sqrt(R^2 + XL^2)

So I = V / Xnet = V / sqrt(R^2 + XL^2) = V / sqrt(R^2 + (2*pi*f*L)^2)

This equation will give you one of the choices.

I hope this helps,

Steve

Physics

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