Can a kind person please check my solution? This is the question: Demonstrate the total current in an RLC parallel circuit using the following circuit. I cant draw it but the inductive reactance is 20ohms, capacitive reactance is 5ohms and resistor of 10ohms. All are in parallel. The voltage is 5v in series with 20ohms inductive reactance.
This is my attempted solution:
Resistive Branch: I = E/R = 5/10 = 0.5A
Capacitive Branch : E/Xc = 5/5 = 1A
Inductive Branch = E/XL =5/20 = 0.25A
Resultant current, Ix = 1 - 0.25 = 0.75a
Total current = (I^2R + I^2X)^1/2
= (0.75^2 + 0.5^2)^1/2 A
Need to see the circuit here.
Dont know how to load it. Sorry. Is the information incomplete?
Ok I've reread and I think I've got it.
Supply + inductive reactance then 3 branches,( one each of R L and C )
There wont be 5v acorss the branches. Some PD will be across the inductive reactance.
You're on the right lines in taking phaes differecnes into account vectorially - but then you need to add in the PD acorss the inductance in the main circuit to get 5v.
Thanks for posting! You just need to create an account in order to submit the post
Already a member?
Oops, something wasn't right
please check the following:
Not got an account?
Sign up now
© Copyright The Student Room 2016 all rights reserved
The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.
Register Number: 04666380 (England and Wales), VAT No. 806 8067 22
Registered Office: International House, Queens Road, Brighton, BN1 3XE