Electrical Theory Help

[R Hutcheson]

I'm attempting to expand my knowledge and grasp of electrical systems. To that end I am reading Boat Owners Illustrated Electrical Handbook, and actually working through the practice problems. I am absolutely stumped by the last two "Q" and "R". Which indicates to me I am failing to grasp a fundamental concept which would allow me to solve these. The answer key has not helped, since it only give the final answer without showing how it is derived.

Is there anyone out there who can explain how to solve these problems?

Thanks

Roy
ROAM

[thomm225]

Pretty simple, what's the question.

(Not trying to put you down. I've been an electronics tech since the mid 70's. We don't get to enjoy it much these days though due to the computer industry and how there are so few failures especially in circuits you can actually replace components in due to the high tech soldering and tiny components)

Current in the legs? Current splits at each intersection but will total the max when it comes back together.

Voltage? In one diagram, the voltage dropped to 9 volts from 12 volts due to the inline resistor.

Always remember I=E/R.

You can figure most things out with that.

And remember current splits.

[wholybee]

I won't do the math for you because I am lazy. But will walk through how to get the answer for Q, same basic technique for R.

First, simplify the diagram to just the voltage source and a single resistance. What is the resistance of 100ohm and 150ohm in parallel? Now take that answer and place it in series with 10ohm. The answer of that, and it is in parallel with 6ohm. The answer to that is in series with 5 ohm. Now, with just a voltage and single resistance, you can find I. Then work backwards from what you just did to find the current in each leg of two parallel resistances. Remember in series resistance the current is the same in both resistances, but in parallel current is divided.

[Bowdrie]

Circuit resistance when resistors are in parallel is less than the value of the lowest resistor.
https://www.khanacademy.org/science/...llel-resistors

[Chotu]

I decided to give this a shot because it had been a while. Namely, decades.

I chose to tackle Problem Q first.

I don’t like where it went. Lol

first, I reduced the 150 ohm and 100 ohm resistors down to a single resistor of 60 ohms

not bad.

then I noticed the 10 ohm resistor and my newly created 60 ohm resistor are in series. So I added those together to get a 70 ohm resistance.

seems good.

then I noticed the six ohm resistor and my new 70 ohm resistor are also in parallel. So I reduced those down. But those didn’t go nicely. So something is wrong.

when I took the 70 ohm resistor and the six om resistor an applied the resistor equivalent law to it,. I got (1/6 + 1/70)^-1. and this evaluates to 5.53? That’s not a good sign.

when people create these problems they make everything even numbers.

from there it should have been very easy because the five ohm resistor would now be in series with the 5.53 equivalent resistor giving you 10.53 ohms total circuit resistance.

then using V equals IR, the current in the circuit near the battery is 12÷10.53 or 1.14 amps.

Which is completely garbage because they never use numbers that aren’t round in these problems.

although I don’t really see where I went wrong either. I just followed this normal protocol.

sort of fun, but seems like I got defeated after 30+ years without practicing one of these.

[Statistical]

Originally Posted by Chotu

I decided to give this a shot because it had been a while. Namely, decades.

I chose to tackle Problem Q first.

I don’t like where it went. Lol

first, I reduced the 150 ohm and 100 ohm resistors down to a single resistor of 60 ohms

not bad.

then I noticed the 10 ohm resistor and my newly created 60 ohm resistor are in series. So I added those together to get a 70 ohm resistance.

seems good.

then I noticed the six ohm resistor and my new 70 ohm resistor are also in parallel. So I reduced those down. But those didn’t go nicely. So something is wrong.

when I took the 70 ohm resistor and the six om resistor an applied the resistor equivalent law to it,. I got (1/6 + 1/70)^-1. and this evaluates to 5.53? That’s not a good sign.

when people create these problems they make everything even numbers.

from there it should have been very easy because the five ohm resistor would now be in series with the 5.53 equivalent resistor giving you 10.53 ohms total circuit resistance.

then using V equals IR, the current in the circuit near the battery is 12÷10.53 or 1.14 amps.

Which is completely garbage because they never use numbers that aren’t round in these problems.

although I don’t really see where I went wrong either. I just followed this normal protocol.

sort of fun, but seems like I got defeated after 30+ years without practicing one of these.

Looks correct to me. Sometimes they use uneven numbers of harder problems.

I think the example the OP needs to find I5 but once he find I from the whole circuit (which you did above) the same logic can be used to find I5.

OP I believe I5 is 0.036A (I4 is 0.054A) and current through the 6 Ohm resistor is 1.05A. which lets us double check because total current is 1.14A = 0.036+0.054+1.05 = 1.14A.

[Statistical]

OP this might help because it walks you through simplifying the resistor network with checks at each step and explanations

https://www.khanacademy.org/science/...istor-networks

[Chotu]

Originally Posted by Statistical

Looks correct to me. Sometimes they use uneven numbers of harder problems.

I think the example the OP needs to find I5 but once he find I from the whole circuit (which you did above) the same logic can be used to find I5.

Thank you for doublechecking my work.

I’m glad it seems correct.

I didn’t proceed to break it out to the I5 value because I figured I'd messed something up with the non-integer current.

now it should be fairly straightforward for him to calculate I5 by working his way back through the circuit

[thomm225]

Thanks for posting this.

Our tech's these days spend so little time on electrical circuits due to all the computer hardware, software, and cyber they need to learn.

Thousands of pages to study for each with a small amount of time spent on old school electronics.

But it's best to learn to troubleshoot.

And that would be hard for me to explain because first you might need a few years of theory then lots of problems to solve to learn best

Just remember though, most of the time electrical problems on a sailboat are rarely due to failed circuits but usually corrosion of some sort so a reseat/cleaning of a connection 9 times out of 10 will fix the problem.

[Chotu]

Originally Posted by thomm225

Just remember though, most of the time electrical problems on a sailboat are rarely due to failed circuits but usually corrosion of some sort so a reseat/cleaning of a connection 9 times out of 10 will fix the problem.

This is sooooo true.

you will never use these LRC diagrams to work on a boat.

and it is usually a connection that’s bad and needs a cleaning

Instead of studying these, it’s better to study how many amps you can get at a certain voltage through a certain gauge wire safely with a good margin of safety. That’s the most frequent thing that comes up

And just adding up loads and figuring out how many days you want to last with certain things on

also getting up to speed on AC electrical waveforms and what’s actually happening on the AC side is pretty important as well. Know your hot, neutral and ground and what to do with them in various situations

Know how to use breakers and fuses to protect wires.

once in my life I was able to use this type of information that you are studying on a boat.

I had a VHF radio from God knows what year. But I was fixing it in the 1990s. Lol

it blew a resistor and I was able to isolate this fact. I was able to go to RadioShack, yes this was a long time ago, and get a new resistor and solder it in. Fixing the VHF.

resistors are now the size of a grain of salt if not just part of an IC chip, so you will not be doing that anymore

[Orin]

This is a good text book if you’re actually interested in getting into the theory enough to be dangerous. Probably not as practical as it may seem, but it’s all fairly simple for most to understand.

[Wotname]

Roy,

Q has been sorted upthread so let's look how to resolve 'R'.

You have been given the total circuit current (2A) and various resistances in the circuit.

You are asked to determine the supply voltage (Vs).

Vs = Circuit current times circuit resistance (basic ohms law)

Therefore you need to determine the circuit resistance from the given resistance values.

1. resolve the parallel 100 ohm and 150 ohm resistors to a single value. These become 60 ohms.

2. Add this to the series 10 ohm which becomes 70 ohms

3. Resolve the parallel 6 ohm and 70 ohm to a single value. These become 5.5 ohms.

4. Add this to the series 5 ohm resistor which becomes 10.5 ohms. This is the circuit resistance.

So Vs = 2 amps x 10.5 ohms

Thus Vs is 21 volts.

If anything is uncertain about the above, please ask!

[R Hutcheson]

Originally Posted by Statistical

Looks correct to me. Sometimes they use uneven numbers of harder problems.

I think the example the OP needs to find I5 but once he find I from the whole circuit (which you did above) the same logic can be used to find I5.

OP I believe I5 is 0.036A (I4 is 0.054A) and current through the 6 Ohm resistor is 1.05A. which lets us double check because total current is 1.14A = 0.036+0.054+1.05 = 1.14A.

Statistical,

You seem to have a firm grasp of the problem. I was able to work the problem through to determine the amperage for the circuit as a whole to be 1.14 amps, just as Chotu did. The problem I am having is I have no idea how to take that information and work back through the circuit to arrive at .036 amps for the leg I5. Your answer is correct according to the key.

How did you arrive there?

Thanks again for your help.

[R Hutcheson]

Thank you everyone. The missing piece was how to calculate the voltage drop over individual resistors. I have it now.

Thanks again.

[Statistical]

Originally Posted by R Hutcheson

Statistical,

You seem to have a firm grasp of the problem. I was able to work the problem through to determine the amperage for the circuit as a whole to be 1.14 amps, just as Chotu did. The problem I am having is I have no idea how to take that information and work back through the circuit to arrive at .036 amps for the leg I5. Your answer is correct according to the key.

How did you arrive there?

Thanks again for your help.

Ok so we know I = 1.14A.

Resistance causes a voltage drop. At each resistor we need to compute the voltage drop in order to determine the subsequent current split.

It really is just repeating two two things: computing voltage drop over a resistor anytime we pass a resistor AND compute current split any place the circuit splits. We just trace a path to the resistor in question (I5)

So first resistor is 5 Ohm and we have 1.14A flowing through through it.
V= I*R
V(drop) = 1.14 A * 5 Ohm.

So the voltage at the point after the first resistor and at the first junction is 12V-5.7V= 6.3V.

The two equivalent resistor values (our simplified paths discussed above) at that first junction are 6 Ohm & 70 Ohm. How much current flows across each portion of the circuit.

I=V/R

center: I = V/R I =6.3V / 6 Ohm = 1.05A (we don't need this but useful to double check later)
right: I = V/R I =6.3V / 70 Ohm = 0.09A

Now on the right hand path we pass a 10 Ohm resistor so there is a another voltage drop. We need to compute the voltage after that resistor to do the last split correctly.

V= I*R
V(drop) = 0.09 A * 10 Ohm.
V(drop) = 0.9V

Voltage after the 10 Ohm resistor = 6.3V - 0.9V = 5.4V

Final split

left: I = V/R I =5.4V / 100 Ohm = 0.054A
right(I5): I = V/R I =5.4V / 150 Ohm = 0.036A
I5 has 0.036A of current.

Double check work:
We know (kirchhoff's law) the sum of the current through all the partial circuits has to equal the sum of the total circuit. 1.05A + 0.054A + 0.036A = 1.14A (the computed total current).

As someone else said it is unlikely you will ever need these kind of puzzles on a boat directly but they are useful in helping to reinforce Ohm's law. It helps you (or at least me) to understand things like voltage drops due to resistance so that is why higher amperage circuits need larger cables. Larger cable has less resistance meaning for same current a lower voltage drop. Likewise longer circuits need larger cables to keep the same voltage drop because cabling resistance isn't static it is resistance per foot/meter.