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RLC parallel circuit total cuurent Watch

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    Hi
    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
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    Need to see the circuit here.
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    (Original post by teachercol)
    Need to see the circuit here.
    Dont know how to load it. Sorry. Is the information incomplete?
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    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.
 
 
 
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