Homework finding values of circuit

[dirtyboy161]

The question reads: Three resistors are connected to a source of 240 volts. The first resistor passes current of 240 mA. The second has a voltage drop of 110 volts and the third one has a resistance of 208 ohms.
Find: a) V1
b) R1
c) R2
d) I2
e) V3
f) RT

So i went under the assumption that it is a series circuit and drew the circuit with a triangle under each resistor and started to fill in the blanks of the triangle with the values given and then got all messed up trying to follow the laws and fill in the blanks.

 

[blueroomelectronics]

Why are you starting another thread with exactly the same question?

 

[dirtyboy161]

Sorry first day on site and did not see the homework section until i had already posted the first time. Apologies not trying to make waves i will learn more as i use more.

 

[blueroomelectronics]

The answers in that thread should give you enough to go on. Did you just want someone to do the work for you?

 

[dirtyboy161]

No i am not asking for someone to give me the answers but to help me arrive at the answers. I am completely new to the school thing 45 yrs. old and can't seem to get the right frame of thinking going.

Hers what i have figured so far.I use the triangle under each resistor on my drawing. The first triangle I have 240v over 240 mA × 1000Ω.

The second triangle i am assuming a voltage drop of 110V which i figure leaves me 130V but the law of a series circuit says voltage stays the same which would be 240v so 240V÷240mA = 1000Ω

The third one i have voltage remains the same again 240V ÷ 208Ω = 1.15A

Does this look right to you?

 

[BrownOut]

The second triangle i am assuming a voltage drop of 110V which i figure leaves me 130V but the law of a series circuit says voltage stays the same which would be 240v so 240V÷.240A = 1000Ω

No! The current stays the same in a series circuit. Use the triangle (aka Ohm's law, which is the correct name ) to calculate the resistance with the 110 vots drop, and you already know the current.

 

[dirtyboy161]

so 110V divided by 240mA =458.3 ohms?

or is it 240V minus the 110V drop which leaves 130V?

V1 = 240v
R1 = 1000 Ohms
R2 = 458.3 Ohms
I2 = 240mA
V3 = 50V
Rt = 1666.3

 

[BrownOut]

or is it 240V minus the 110V drop which leaves 130V?

Who would you think the voltage that's left over is important? The triangle says the voltage drop across the resistance, current thru the resistance...

EDIT: I was confused when I wrote this post. I was trying to answer this on two different threads. The votage was not given for the first resistor, and so you do have to calculate it. I think the post below gives this value.

 

[colin55]

This is a very simple problem.
Firstly, you are told the current through one of the resistors in series is 240mA. This is the current though all the resistors because they are in series. By Ohms law, the total resistance will be 240/.240 = 1,000 ohms.

The third resistor is 208 ohms and the current flowing though it is .24A This means the voltage across it is .24 x 208 = 50v
The voltage across the first resistor is 240 - 110 - 50 = 80v
The resistance of R1 = 333R
The resistance of the second resistor = 1,000 - 208 - 333 = 459R

 

[dirtyboy161]

I was getting every thing right but i was not understanding the voltage i was starting out with 240V ÷ .24A = 1000Ω. I wasn't getting the formula 110V+50V=80V for the first resistor. "applied voltage is used by the circuit when in series"

Thanks for making me understand I really appreciate it!

 

[Mikebits]

No i am not asking for someone to give me the answers but to help me arrive at the answers. I am completely new to the school thing 45 yrs. old and can't seem to get the right frame of thinking going.

Hers what i have figured so far.I use the triangle under each resistor on my drawing. The first triangle I have 240v over 240 mA × 1000Ω.

The second triangle i am assuming a voltage drop of 110V which i figure leaves me 130V but the law of a series circuit says voltage stays the same which would be 240v so 240V÷240mA = 1000Ω

The third one i have voltage remains the same again 240V ÷ 208Ω = 1.15A

Does this look right to you?

Here is a problem and a detriment to you. Relying on a crutch like the triangle will only hinder you in progressing further in any science field. All you need is a little simple algebra and remember the basic ohms law equation E=I*R, to derive any parameter. Remember what ever you do on one side of the equal sign you do to the other side. So to isolate I for example; we divide both sides by R and the we have I=E/R.

I was once like you, and my math was appalling and I carried a little triangle pictogram in my wallet so that I would not be caught off guard. I was lost without this little card and would flounder without it.

What a relief it was when I finally learned algebra and no longer needed the sacred triangle. A little confession here, I struggled horribly with math, I had to take college algebra 3 times before I passed with a mere B. Not all of us mere mortals are good at math and for some, such as myself it is a hard and trying path, but getting to the other side of the tunnel brings much satisfaction and a greater understanding in a great many things.

Just my thoughts... Good luck with your problem.

 

[RCinFLA]

Can have two solutions depending on how literal you want to obey Engish.

One solution could be all three resistor in series, but middle resistor has neither of its leads connected to 240 v source.

Other solution is R1 is across 240v and R2 and R3 are in series and across 240v source. This obeys strick English interpretation.