Student principles question help!

[Jonathan Harris]

Hello all

I need help on a question I have from my Level 2.

A 17.46μF capacitor is connected in series to a resistance of 200Ω. It is connected to an ac 230v supply at 50Hz. Calculate:

a. The impedance
b. The current flow

I have:

a. √200²+17.46×10−6²= 200Ω
b. 230/200=1.15A

Is this correct?

Or do I use my capacitive reactance to figure out the current for b?

Thanks!

 

[Lucien Nunes]

√200²+17.46×10−6²

Oopsie! You've vectorially added the capacitance to the resistance, instead of the capacitive reactance.
First, obtain Xc from C. Calculate Z, then use V/Z as you have done to get I.

 

[telectrix]

Xc = 1/ (2pi x f x C), work that out and add to the 200. then I=V/Z

 

[Jonathan Harris]

I see!

So:

a. √200²+182²= 270Ω
b. 230/270 = 0.85Α

Correct?

 

[happysteve]

Good lad.

 

[Jonathan Harris]

Awesome!

Thanks everyone

 

[Lucien Nunes]

add to the 200

And while you're doing that, in quadrature as per your OP, don't forget to write in the brackets! We knew what you meant but as typed the square root only applies to the 200² because addition takes lowest priority.

i.e. √(200²+182²)= 270Ω

BTW, I wonder where the question setter gets his 17.46μF capacitors...

 

[Besoeker]

Hello all

I need help on a question I have from my Level 2.

A 17.46μF capacitor is connected in series to a resistance of 200Ω. It is connected to an ac 230v supply at 50Hz. Calculate:

a. The impedance
b. The current flow

I have:

a. √200²+17.46×10−6²= 200Ω
b. 230/200=1.15A

Is this correct?

Or do I use my capacitive reactance to figure out the current for b?

Thanks!

Your answer (a) implies that the capacitor adds no impedance to the circuit. Do you think that's likely?

Oops. Late again.

 

[hightower]

(Ignore me, not reading replies correctly)

 

[Besoeker]

BTW, I wonder where the question setter gets his 17.46μF capacitors...

Same place he gets his wine?

 

[Lucien Nunes]

Your answer (a) implies that the capacitor adds no impedance to the circuit. Do you think that's likely?

Yes, that's a good double-check. You're unlikely to be asked questions where one of the supplied values has so little effect on the result that it is lost in the noise. It happens a lot in the real world (e.g. a 20mΩ busbar has no tangible effect on the leakage from 75MΩ insulation) but it would just be a timewaste in a set question.

 

[DuaneMHunt1976]

dont forget your brackets
Xc = 1 / (2 x Pi x f x C)
Xc = 1 / (2 x 3.14 x 50 x (17.46×10⁻⁶))
Xc = 1 / 0.00548244
Xc = 182.4005369872


Z = √ (R² + (Xc-XL)² )
Z = √ ( 200² + 182.4005369872² )
Z = √ ( 40000 + 33269.9558932119 )
Z = √ 73269.9558932119
Z = 270.684236507

270.7 Ohms to 1dp

I = U / Z
I = 230 / 270.684236507
I = 0.849698538

0.85A or 850mA

 

[HandySparks]

dont forget your brackets
Z = √ 73269.9558932119
...
I = 0.849698538

0.85A or 850mA

Little point in showing numbers to 15 significant figures if you're going to round off to 2 in the end. 4 or 5 will be enough, especially as you've used pi to only 3 sig figs.

 

[DuaneMHunt1976]

Pi = C/d
C= ((360/x)(SinX/H))
Were X is a dam small angle and H is best to be called 1 to make the Maths simple

Pi = ((360/x)(SinX/H))/2H
Pi = ((360/x)(SinX/1))/2

sadly i cant remember Pi to too many dp
ive never liked rounding up or down as i feel it isnt a true rep of the value you are after.

 

[Richard Burns]

3.141592653589793238462643383279502284197169399275820 is as far as i remember it, I have forgotten some digits over time.:rolleyes4:

 

[Besoeker]

Pi = C/d
C= ((360/x)(SinX/H))
Were X is a dam small angle and H is best to be called 1 to make the Maths simple

Pi = ((360/x)(SinX/H))/2H
Pi = ((360/x)(SinX/1))/2

sadly i cant remember Pi to too many dp
ive never liked rounding up or down as i feel it isnt a true rep of the value you are after.

Rounding is real life. What if the resistor in the OP has a tolerance of 5% ?
I know it doesn't state that so the test question has to be answered in the terms given. And to the resolution of the stated values.