# Electronics RLC Ciruit question

Discussion in 'Homework Help' started by jeffrey1027, Nov 9, 2014.

1. ### jeffrey1027 Thread Starter New Member

Nov 9, 2014
5
0

I started by finding
ZL = jwL which gives j(672 Ohms)

Don't know where to go from here.

TIA

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,963
615
Then you find voltage across your 672j Ohm "resistor".

Also. RLC circuits are not considered electronics.

jeffrey1027 likes this.
3. ### jeffrey1027 Thread Starter New Member

Nov 9, 2014
5
0
Would it be
VL = IL
=( - 65 x 10^-3 A) (672j)
= -j 43.68V

Would I just add that to the Given Vs?
Vx = Vs + VL

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,963
615
Yes.
But. Now you need to watch out your polarity.

VX=VS+VL That is correct. But if your polarity is wrong, you will use VL with wrong sign and get wrong answer.

jeffrey1027 likes this.
5. ### jeffrey1027 Thread Starter New Member

Nov 9, 2014
5
0
Would it not just be :
Vx = (-55 + j22 V) + ( 0 - j 43.68 V)
= -55 - 21.7j V
?

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,963
615
Replace the inductor with a voltage source. Place the polarity symbols into the voltage source. Now you have two voltage soruces in series. Do you know how to "add" two voltage sources in series?

jeffrey1027 likes this.
7. ### jeffrey1027 Thread Starter New Member

Nov 9, 2014
5
0
I think i figured out where the mistake I made
Vx = -VL + (-Vs)
= - ( 0 - j 43.68 V) + [- (-55 + j22 V)]
= + 55 + 21.7j V

Isn't adding voltage sources the same as adding resistors in series? Req = R1 + R2 +...+ Rn

8. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,963
615
The current enters the "resistor" from the bottom and moves up. The terminal where current enters is assumed to be positive. The terminal where current exits the "resistor" is assumed to be negative. The current flows from higher voltage potential to lower voltage potentia.

So the bottom terminal is positive, upper terminal is negative. But the voltage across the "resistor" is -43.68j volts. What does that mean? It means that our earlier assumptions about which terminal is at higher electrical potential (positive) were wrong. So we will do a simple thing. We will make upper terminal of the "resistor" positive, bottom terminal will be negative, and now instead of -43.68j volts we have +43.68j volts.

Also now we have two voltages connected from positive pole to negative pole, now we really can add two series voltages: Vx=-55+22j+43.68j

jeffrey1027 likes this.
9. ### jeffrey1027 Thread Starter New Member

Nov 9, 2014
5
0
Thanks a lot for the help!