# RLC (Impedance) Series/Parallel Question

#### johncena

Joined Aug 15, 2010
32
Here's the prob.

My progress so far is:

i converted V(t) to: v(t) = 120cos(377t -30°) just minues 90° to make it cos

then frequency is
F = 377/ 2∏
F = 60
XL = 2.pi(60)(40mH)
XL = j15.08 or (0+j15.08)

XC = - [1/(2∏(60)(50uF))]
XC = -j53.05 or 0-j53.05

XLR=20 + j15.08

so this is the part i am confused...
We need to parallel XLR and XC so the equation is this

ZT = (XLR)(XC) / (XLR + XC)

is this correct? or i need to convert it to polar form? but i get infinity at :
Φ = tan^-1 (53.05 / 0 )

How to solve this?
i can solve i(t) by the formula i(t) = V/ZT
but i can't solve ZT.

#### Georacer

Joined Nov 25, 2009
5,154
Eiter method, polar or cartesia representation must yield the same result. I would say, use cartesian form to find the sum on the denominator and polar to find the product on the nominator. Then polar again to find the ratio.

The conversions can be done very easilly with an adequate pocket calculator.

#### johncena

Joined Aug 15, 2010
32
is this correct?
XC = - [1/(2∏(60)(50uF))]
XC = -j53.05 or 0-j53.05

is it negative? because the is formula positive to some books/websites.
and.. if it is -j53.05, the polar form will be (-53.05<-90°)???
(Note: < means right angle)

#### johncena

Joined Aug 15, 2010
32
Is Xc is negative based on the formula? because the formula is positive to some books/websites.

Our given formula was Xc = -[ 1/(2∏fC)]

And..

If i converted -j53.05 to polar form.. is this correct? -> -53.05<-90°, or not?
(note: < means right angle)

#### johncena

Joined Aug 15, 2010
32
Modified the Image: It was Dc not AC >.<

Is Xc is negative based on the formula? because the formula is positive to some books/websites.

Our given formula was Xc = -[ 1/(2∏fC)]

And..

If i converted -j53.05 to polar form.. is this correct? -> -53.05<-90°, or not?
(note: < means right angle)

Last edited:

#### Georacer

Joined Nov 25, 2009
5,154
Actually it is $X_c=\frac{1}{j \omega C}=-j\frac{1}{\omega C}=\frac{1}{\omega C}\angle -90$

#### bertus

Joined Apr 5, 2008
20,049
Hello,

I noticed two posts in the background.
I made them visible for you.

Bertus

#### Georacer

Joined Nov 25, 2009
5,154
-j53.05 is correct, but it is equal to $53,05\angle-90$, not $-53,05\angle-90$.

#### johncena

Joined Aug 15, 2010
32
-j53.05 is correct, but it is equal to $53,05\angle-90$, not $-53,05\angle-90$.
ah now i see... thanks

PS: If by any chance i have a value of XC whiich is postive... ex. +j40.. the polar form will always with -90°? like this.. --> 40<-90°

#### blah2222

Joined May 3, 2010
582
Modified the Image: It was Dc not AC >.<
V(t) is a time-varying system, so this is still steady-state AC. The Symbol for the voltage source was fine before.

#### Georacer

Joined Nov 25, 2009
5,154
Capacitors are always leading, so their angle is always negative. But I 'll generalize the topic for any imaginary number.

Let's suppose we have an imaginary number X=A+jB, A and B are either positive or negative. This is a cartesian representation in a rectangular grid.
If we wanted to turn this number into polar coordinates we would write $X=M\angle \theta$ where $M=\sqrt{A^2+B^2}$ and $\theta=\arctan\left( \frac{B}{A} \right),\ -90^o\leq \theta \leq 90^o$.
It is also equivalent to write $X=-M\angle -\theta$ but the convention is to have the M positive.