Hi,

I have this simple circuit

i Just need to find the phasor current I. But i must be doing something wrong,
because my answer differs a bit from the answer in the book.

1. Replace the capacitance with it's complex impedance
\(-\frac{1}{333\cdot10^{-5}} = -J300.3\)

2. find \(Z_eq = \frac{1}{1/115 + 1/-300.3} = 107.39 \)

3. 1585/107.39 = 14.76, the correct answer is supposed to be 15.11.

What am i doing wrong?

 

You forgot to include the j in the Zeq equation.

 

You forgot to include the j in the Zeq equation.

How do you mean? The angle?

 

It should be:

Zeq=1/[1/115-1/j300.3]=1/[1/115+j/300.3]

which equals:

Zeq=1/(0.0093<20.95)=107.53<-20.95

 

On the picture the voltage source will be 1585 * 1.4142 (root 2) i.e 2241.19 Volt and the polar angle is 12 deg.

Good luck

 

Hmm, I'm getting the same answer as the OP.

 

On the picture the voltage source will be 1585 * 1.4142 (root 2) i.e 2241.19 Volt and the polar angle is 12 deg.

Good luck

But you still use 1585 rms for the calculation right?

 

Accurately,

Zeq=107.39<-20.95 Ω

and if the RMS value for voltage is used,

I=14.759<32.95 A

 

What am i doing wrong?

 

What am i doing wrong?

These answers are wrong!

 

Ok, thanks for clearing that up.
I f#&%? hate my book.

 

Ok, thanks for clearing that up.
I f#&%? hate my book.

Well, do not always trust me too.

 

Hi,

I have one last question.

If i have parallel RLC phasors, how do i get the angle of the resulting phasor.

For example, i would like to find the equivalent of these two parallel phasors \( 37.94\angle71.5\) and
\( 133.33\angle-90\)

I know how to get |Z| , just not how to get the angle.

\(\frac{1}{\frac{1}{37.94} + \frac{1}{133.33}} = 29.53\)