ok, thanks for the reply, to my last question. Here is another one, which I can't figure out. Here is the situation a circuit consist of the following: 18 volt source, the circuit is a series-parallel set up like this: voltage source to r1 at 2 ohms, r2 at 2 ohms then the circuit turns to a parallel circuit r3 at 4 ohms r4 at 4 ohms then back to series with r4 at 2 ohms. Here is the question what is the volt meter reading between r3? I looked up the answer in the book and they have 1.5 volts, also they have 0.75 amps for this circuit, I just can't figure out how they came up with 0.75 amps. If I did this right the total resistance for this circuit should be about 8 ohms, but then again what do I know. any help would be appreciated
resistors
- Thread starter rudy.v
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bphgravity
Senior Member
- Location
- Florida
Re: resistors
If I read your circuit description right, you are saying that two 4-ohm resistors are in series with three 2-ohm loads. In that case, start by reducing the parallel loads.
4x4/4+4=2-ohm, Now you have four 2-ohm resistors in series. 2 x 4 = 8-ohms total. 18 / 8 = 2.25-amps.
The first two resistors drop the voltage 9 volts. 2-ohm x 2.25-amp = 4.5-volt. This puts 9-volts at both R3 and R4. This leaves a remainder of 4.5-volts at the last resistor. 4.5 + 9 + 4.5 = 18-volts.
I don't know where the 1.5-volt and .75-amp is coming from unless I am drwing the circuit wrong?
If I read your circuit description right, you are saying that two 4-ohm resistors are in series with three 2-ohm loads. In that case, start by reducing the parallel loads.
4x4/4+4=2-ohm, Now you have four 2-ohm resistors in series. 2 x 4 = 8-ohms total. 18 / 8 = 2.25-amps.
The first two resistors drop the voltage 9 volts. 2-ohm x 2.25-amp = 4.5-volt. This puts 9-volts at both R3 and R4. This leaves a remainder of 4.5-volts at the last resistor. 4.5 + 9 + 4.5 = 18-volts.
I don't know where the 1.5-volt and .75-amp is coming from unless I am drwing the circuit wrong?
mikeackley
Senior Member
- Location
- Washington
Re: resistors
Rudy: I'm having some problems following your description of this circuit.
If I diagram what I think you've described, it would look like this.
[ April 16, 2003, 09:41 PM: Message edited by: mikeackley ]
Rudy: I'm having some problems following your description of this circuit.
If I diagram what I think you've described, it would look like this.
[ April 16, 2003, 09:41 PM: Message edited by: mikeackley ]
Re: resistors
Originally posted by bphgravity:
If I read your circuit description right, you are saying that two 4-ohm resistors are in series with three 2-ohm loads. In that case, start by reducing the parallel loads.
4x4/4+4=2-ohm, Now you have four 2-ohm resistors in series. 2 x 4 = 8-ohms total. 18 / 8 = 2.25-amps.
The first two resistors drop the voltage 9 volts. 2-ohm x 2.25-amp = 4.5-volt. This puts 9-volts at both R3 and R4. This leaves a remainder of 4.5-volts at the last resistor. 4.5 + 9 + 4.5 = 18-volts.
I don't know where the 1.5-volt and .75-amp is coming from unless I am drwing the circuit wrong?
Re: resistors
ok, Mike you have the circuit diagram correct the voltage question is between r3 & r2 & r3 & r5.
ok, bphgravity I came up with the same numbers you came up with, that is why the answer that is in the book ( 1.5 v & .75 amps) did not make sense to me, the 1.5 and .75 were numbers provided by the book I am currently studing, which happens to be Mike Holt's Illustrated Guide Electrical Exam Preparation, maybe it is just an typo error. Thank you all for your responces. it dies help.
ok, Mike you have the circuit diagram correct the voltage question is between r3 & r2 & r3 & r5.
ok, bphgravity I came up with the same numbers you came up with, that is why the answer that is in the book ( 1.5 v & .75 amps) did not make sense to me, the 1.5 and .75 were numbers provided by the book I am currently studing, which happens to be Mike Holt's Illustrated Guide Electrical Exam Preparation, maybe it is just an typo error. Thank you all for your responces. it dies help.
mikeackley
Senior Member
- Location
- Washington
Re: resistors
Well...shame on Mike. Guess his technical writers need to spend more time right here in River City...like on this Forum, learning with the rest of us.
Well...shame on Mike. Guess his technical writers need to spend more time right here in River City...like on this Forum, learning with the rest of us.
Re: resistors
Well maybe it was an off day for them, anyway Mike like I said you have the diagram correct what do you think of this problem ? by the way how did you do the diagram on the computer (I am not as computer literate as I would like to be, but I am always learning)again thanksOriginally posted by mikeackley:
Well...shame on Mike. Guess his technical writers need to spend more time right here in River City...like on this Forum, learning with the rest of us.
mikeackley
Senior Member
- Location
- Washington
Re: resistors
Rudy:
I don?t have a lot of fancy software either, so how I do simple diagrams like the one above is a somewhat cumbersome multi step process.
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Not a pretty process, but it works.
Rudy:
I don?t have a lot of fancy software either, so how I do simple diagrams like the one above is a somewhat cumbersome multi step process.
</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">First I bring up MS WORD and draw whatever. (You could use EXCEL or PowerPoint just as well)</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Next I open up a Photo Editor program [my computer just happened to come with Adobe PhotoShop already loaded, but there are a ton of other ones out there] and I do a Copy of the WORD drawing, and Paste onto a new Picture created by Adobe PhotoShop. Then save this ?picture? as either a GIF or JPEG file.</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">Then I upload the GIF or JPEG file to my Homepage file space provided by my Internet Service Provider (earthlink) where it is available to the world, with the proper URL that is.</font>
Not a pretty process, but it works.
mikeackley
Senior Member
- Location
- Washington
Re: resistors
Rudy: My numbers would be the same as what Bryan came up with.
[Edit]: P.S. -- Or to be more specific, Voltage between R2 & R3 = 9V. Then, between the parallel pair (R3 & R4) and R5, voltage would be 4.5V. Is that what you needed rudy?
[ April 17, 2003, 02:10 PM: Message edited by: mikeackley ]
Rudy: My numbers would be the same as what Bryan came up with.
[Edit]: P.S. -- Or to be more specific, Voltage between R2 & R3 = 9V. Then, between the parallel pair (R3 & R4) and R5, voltage would be 4.5V. Is that what you needed rudy?
[ April 17, 2003, 02:10 PM: Message edited by: mikeackley ]
Re: resistors
Mike: yes this is what I was looking for I thought I had the right numbers,but the answers in the book were throwing me off. I guess the voltage between R1 would be 4.5, and then adding the voltage between R 2 & R3 9 volts, and R3,R4 & R5 4,5 volts comes out to the voltage source of 18 volts? Thanks for your responce and help.
mikeackley
Senior Member
- Location
- Washington
Re: resistors
Oops. Voltage between R1 & R2 is 13.5V. The best of luck Rudy with the exam.
[ April 18, 2003, 01:54 AM: Message edited by: mikeackley ]
Oops. Voltage between R1 & R2 is 13.5V. The best of luck Rudy with the exam.
[ April 18, 2003, 01:54 AM: Message edited by: mikeackley ]
Re: resistors
Mike: Thanks to you and everyone else for the help on these problems, I am just getting started in this field, but it's good to know that there are individuals out there that still care about helping those who need assistance. I hope I don't get on any one's nerves with all my questions, but I have all ways said when in doubt ask someone who knows, and it seems to me you guys know.Originally posted by mikeackley:
Yep, you've got it right. Voltage between R1 & R2 is 4.5V. The best of luck Rudy with the exam.
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