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Can someone show me where im going wrong please?

A 0.25H inductor has a resistance of 35 Ohms and is connected in series with a capacitor of 42uF. they are connected to an AC 230V supply at 50Hz, what is the circuits current? show your working ?
 
Although the resistance is stated to be the DC resistance of the inductor, it can be treated as a separate component for analysis. Therefore you could begin with an equation for the impedance of a series LCR circuit.
 
Give us your thoughts and someone might help.
I was thinking:

Inductive reactance XL=2xPixfxL
XL=2x3.14x50x0.25
XL =78.5398 ohms

Capacitive reactance XC=1 / (2xPixfxC)
XC=1 / (2x3.14x50x42uf)
XC=7.578 ohms

Together with the 35 ohm resistor that makes a total resistance of 121.117 ohms

I=V/R
I=230/121.117
I=1.898A

Am I missing anything here?
 
I was thinking:

Inductive reactance XL=2xPixfxL
XL=2x3.14x50x0.25
XL =78.5398 ohms

Capacitive reactance XC=1 / (2xPixfxC)
XC=1 / (2x3.14x50x42uf)
XC=7.578 ohms

Together with the 35 ohm resistor that makes a total resistance of 121.117 ohms

I=V/R
I=230/121.117
I=1.898A

Am I missing anything here

I was thinking:

Inductive reactance XL=2xPixfxL
XL=2x3.14x50x0.25
XL =78.5398 ohms

Capacitive reactance XC=1 / (2xPixfxC)
XC=1 / (2x3.14x50x42uf)
XC=7.578 ohms

Together with the 35 ohm resistor that makes a total resistance of 121.117 ohms

I=V/R
I=230/121.117
I=1.898A

Am I missing anything here?
ok I think i know what i did wrong, im using 42 micro farads in a formal meant to farads ?

so now id say:
Capacitive reactance XC=1 / (2xPixfxC)
XC=1 / (2x3.14x50x42uf)
XC=75.78 ohms
Bringing the total resistance upto 189.32 (189.3198 but you know...)

I=V/R
I=230/189.32
I=1.214A

id greatly appreciate if anyone could double check this for me
 
Am I missing anything here?

Yes. One cannot simply add reactances to resistances as scalar values, because their impedance components are not in-phase. For example, if you pass a current through a resistor and inductor in series, although the current through the two will be in phase, the voltages developed across them will be in quadrature (differing in phase by 90° i.e. Π/2 rad)

Start out with the formula for a series RLC circuit. Quote it here and try either explaining why it is like it is, or even just try plugging your numbers into it and see what comes out. We can explain if you get stuck.

BTW Are you familiar with phasor diagrams and/or complex number notation for AC circuit analysis?
 
Yes. One cannot simply add reactances to resistances as scalar values, because their impedance components are not in-phase. For example, if you pass a current through a resistor and inductor in series, although the current through the two will be in phase, the voltages developed across them will be in quadrature (90° or Π/2 rad apart.)

Start out with the formula for a series LCR circuit. Quote it here and try either explaining why it is like it is, or even just try plugging your numbers into it and see what comes out. We can explain if you get stuck.

BTW Are you familiar with phasor diagrams and/or complex number notation for AC circuit analysis?
so my answer is wrong then? lol and ye iv used phasor diagrams before
 
Yes, it's wrong. You've added the magnitudes of R, Xc and Xl without considering their relative phase angles. You must treat them as vectors, but the actual working can be simplified because Xc and Xl are parallel (although in antiphase) and perpendicular to R. So you can calculate it as a simple triangle.
If you start out with the formula like I suggested, that aspect will become clear.
 
Yes, it's wrong. You've added the magnitudes of R, Xc and Xl without considering their relative phase angles. You must treat them as vectors, but the actual working can be simplified because Xc and Xl are parallel (although in antiphase) and perpendicular to R. So you can calculate it as a simple triangle.
If you start out with the formula like I suggested, that aspect will become clear.
thankyou, i will give it another go now
 
Yes, it's wrong. You've added the magnitudes of R, Xc and Xl without considering their relative phase angles. You must treat them as vectors, but the actual working can be simplified because Xc and Xl are parallel (although in antiphase) and perpendicular to R. So you can calculate it as a simple triangle.
If you start out with the formula like I suggested, that aspect will become clear.
Z=R2+(XL-XC)2 = 7.5625ohms

I=230/7.5625
I= 30.413A
 
Yes, it's wrong. You've added the magnitudes of R, Xc and Xl without considering their relative phase angles. You must treat them as vectors, but the actual working can be simplified because Xc and Xl are parallel (although in antiphase) and perpendicular to R. So you can calculate it as a simple triangle.
If you start out with the formula like I suggested, that aspect will become clear.
right! lets do this again lol

z=the square root of R2+(XL-XC)2
= 35.1078ohms

I=V/R
=230/35.1078
=6.55A
 
Yup, looks right.
Note that because of the particular values of inductance and capacitance chosen in the question, at 50Hz their reactances are almost equal in magnitude and almost cancel out due to being in antiphase.
When Xc = Xl the circuit is at resonance, the impedance is simply the resistance and φ=0.
 
Yup, looks right.
Note that because of the particular values of inductance and capacitance chosen in the question, at 50Hz their reactances are almost equal in magnitude and almost cancel out due to being in antiphase.
When Xc = Xl the circuit is at resonance, the impedance is simply the resistance and φ=0.
brilliant thankyou, could you look at my other post by any chance?
 

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