How to Calculate Currents in an A.C Network?

[Emzielou83]

Hi,

This is a coursework question from my Eng science.

An A.C network consists of a coil of inductance 10mH and series resistance of 25 ohms in parallel with a 50*10-6F capacitor. The A.C supply voltage is 120/0 degrees V at 400Hz.

Draw a well labelled diagram
a) Calculate the current in the capacitor in polar form
b) Calculate the current in the coil in polar form
c) calculate the total current from the supply voltage in polar form.

I have all the final answers:
a) Ic= 15.1Arg 90 degrees Amps
b) ILR = 3.4Arg-45 degrees Amps
c) I = 12.9Arg 79.3 degrees Amps

I am using the formula

Ic = V/Xc
Xc = 1/2piFC

Xc works out to be 7.957 ohms

Then using Ic = V/Xc

Ic = 15.08 Amps

15.08Arg90 degrees.

the next bit is where I get stuck.

For IL I am using

IL = V/XL
XL = 2piFL

I am working XL out at 25.13 ohms

Then IL =V/XL = 4.77 Amps.

I know this is wrong but I can't see what I am doing.

Also I am not too brilliant at working out the polar side of things, if anyone knows a simple way of explaining this it would be a great help.

Thanks

Emma

 

[CEL]

Hi,

This is a coursework question from my Eng science.

An A.C network consists of a coil of inductance 10mH and series resistance of 25 ohms in parallel with a 50*10-6F capacitor. The A.C supply voltage is 120/0 degrees V at 400Hz.

Draw a well labelled diagram
a) Calculate the current in the capacitor in polar form
b) Calculate the current in the coil in polar form
c) calculate the total current from the supply voltage in polar form.

I have all the final answers:
a) Ic= 15.1Arg 90 degrees Amps
b) ILR = 3.4Arg-45 degrees Amps
c) I = 12.9Arg 79.3 degrees Amps

I am using the formula

Ic = V/Xc
Xc = 1/2piFC

Xc works out to be 7.957 ohms

Then using Ic = V/Xc

Ic = 15.08 Amps

15.08Arg90 degrees.

the next bit is where I get stuck.

For IL I am using

IL = V/XL
XL = 2piFL

I am working XL out at 25.13 ohms

Then IL =V/XL = 4.77 Amps.

I know this is wrong but I can't see what I am doing.

Also I am not too brilliant at working out the polar side of things, if anyone knows a simple way of explaining this it would be a great help.

Thanks

Emma

The source voltage is applied to the series connection of the inductance and the resistance: XL + R.

 

[Mr.Green]

IL = V/XL

This is wrong. Since you have a coil of inductance and resistance so, the current is given by:

I_L = V/(R + j X_L)
I_L = 120 < 0^O / (25 + j 25.13) = 3.385 < - 45.148^O

if anyone knows a simple way of explaining this it would be a great help

Any quantity that has a magnitude and a direction can be represented in two ways:

1) Rectangular form; for example:

R + j X_L

is written in rectangular form.

2) Polar form; for example let's convert [R + j X_L] to polar form:

The magnitude is given by:
[ R² + X_L² ]^0.5
and the angle (sometimes also called argument):
tan^-1 [X_L/R]