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?