# RL parallel circuits

Discussion in 'Electronic Basics' started by conrad, May 4, 2007.

I have an RL parallel circuit.
The AC voltage is 120 and the frequency
is 60 Hz. One branch has a resistor (60 ohms)
and the other branch(the one parallel to
the resistor) has an inductor with 60 mH.

The inductive reactance of my inductor is
22.61(rounded to nearest hundreth) ohms.

My question is this, to find the impedance
I use Z = 1/(sqrt([1/60]^2 + 1/[22.61]^2)) ?

2. ### Rich GriseGuest

If that's what the textbook says, then, obviously, yes.

Good Luck!
Rich

3. ### Bill BowdenGuest

I don't do it that way. First I decide what's reasonable, and since
you have a 60 ohm resistor across the source, the total impedance must
be less than 60 ohms. The currents are 90 degrees out of phase, so the
solution is to vector the resistor current at 90 degrees from the
inductor current, get the result, and then convert that to impedance
at some angle.

The resistor current will be 120/60 = 2 amps, and the inductor current
will be 120/(6.28*60*.06) = 5.31 amps. The total current will be
sqrt(5.31^2 + 2^2) = 5.67 amps. Therefore, the impedance will be
120/5.67 = 21.15 ohms. And you can figure out the phase angle.

-Bill