# Admittance Magnitude in a Parallel RL Circuit

Discussion in 'Electronics Homework Help' started by Cine, Sep 28, 2010.

1. ### Cine

1
0
Sep 28, 2010
Hi,

I some have problems regarding the parallel RL circuit and it is the admittance magnitude. I have tried to apply the solution that was given to me in the text book. Here is my attempt on the question:

"R=2 Ohms, L=19 H & e is a 300 V, 18Hz power supply. Calculate the magnitude of the admittance of the circuit?"

The solution:

6.28x19 H x 18Hz=214.88

1/2=0.5

1/214.88=0.046

0.5^2+0.046^2 square root=502

I'm really lost. Thanks.

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2. ### Militoy

180
0
Aug 24, 2010
First step - check the decimal placement on your calculation of inductive reactance.

BTW - the methodology used in the rest looks OK to me. If your numbers above are accurate - you'll note that the inductor makes very little difference to the operation of the circuit - especially after the "first step" above.

Last edited: Sep 28, 2010  