Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Admittance Magnitude in a Parallel RL Circuit

Status
Not open for further replies.

Terra_Ryzing

Newbie level 2
Joined
Sep 24, 2010
Messages
2
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,295
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 need someone to correct me on this. Thanks.
 

Attachments

  • ac_rl_parallel_sch.gif
    ac_rl_parallel_sch.gif
    580 bytes · Views: 125

6.28x19 H x 18Hz=214.88

0.5^2+0.046^2 square root=502
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top