# Series / Parallel circuit question

#### Gdrumm

Joined Aug 29, 2008
684
I'm trying to solve a series / parallel circuit problem, and I don't know where to start. I've tried collapsing R10, R9, and then joining it with R11, but I can't read it well enough yet to determine IT, and I am getting more confused by the minute.

I want to solve it myself, but I really need some solid clues as to where I should start. So many pieces are missing...

I think R7 will be ignored because of the open switch?

Thanks,
Gary

#### guitarguy12387

Joined Apr 10, 2008
359
What are you solving for?

If you use ohms law at the top (after combining R2, R3, R4 in series), you can find the current through R2, etc.

And correct, R7 drops no voltage cuz no current flows through it (because of the open).

So you will then know the current through R8. Then, simple application of ohm's law will give you the currents through the bottom network.

You know most of the stuff you need... mostly ohm's law will do the trick.

#### Gdrumm

Joined Aug 29, 2008
684
Ok, I solved some of it, but I'm lost at R6, R8, and R9.

I don't know if at the top, 2.63A = IT

SInce R2,3,4 and R5,6 have 1.2 A, and 1.42 A respectively (=2.63A)

I don't know how to work my way back, and I think I have to solve for something on the negative side, prior to R11.

Thanks,
Gary

#### studiot

Joined Nov 9, 2007
5,002
The secret is to realise the 2.63A current splits and recombines in several places. Each time you are told the voltage across a parallel combination and the current in one branch of the split, so you can easily work out the current in the other branch and thus the unknown resistor.

Then you can reduce the parallel combination to a single resistor.

For example there is 4.56 volts across the parallel combination of R9+R10 // R11, this plus the current should allow you to work out R9 and so reduce the combination to a single resistor.

Similarly there is 21.7 volts across The series combination of R5 and R6. You are told the current so you can work out the voltage across R5 and by subtraction the voltage across R6. Thus you know the current and voltage for R6 you can work out its value.

Hope this helps.

Last edited:

#### Gdrumm

Joined Aug 29, 2008
684
I'm off by a fraction, and can't find the error.

Does anyone feel like checking my work?

et 50 v
rt 18.8 ohms
it 2.63 a
ir1 1.2 a
ir3 1.2 a
ir4 1.2 a
ir7 0 a
ir8 2.63 a
ir9 912 mA
ir10 912 mA
ir11 2.28 a
er1 10.5 v
er2 9.7 v
er3 7.2 v
er5 7.15 v
er6 14.6 v
er7 13.2 v
er8 13.2 v
er9 2.81 v
er10 1.75 v
er11 4.56 v
r2 8.1 ohms
r6 10.1 ohms
r8 5 ohms
r9 2.56 ohms

Thanks,
Gary

#### The Electrician

Joined Oct 9, 2007
2,736
Does anyone feel like checking my work?

et 50 v
rt 18.8 ohms
..If rt is the total load on the battery, then its value is
..et/it = 50/2.63 = 19.0114
it 2.63 a
ir1 1.2 a
..Shouldn't this be ir2? We already know that ir1 = 2.63 amps.
ir3 1.2 a
ir4 1.2 a
ir7 0 a
ir8 2.63 a
ir9 912 mA
..ir9 = .35 amps. See below.
ir10 912 mA
..ir10 = .35 amps.
ir11 2.28 a
er1 10.5 v
..If R1 is carrying 2.63 amps, and its value is 4 ohms, isn't er1 = 10.52 volts?
..If you're getting results that are off by a fraction, you'd better carry more
..than 3 digits in your calculations.
er2 9.7 v
er3 7.2 v
er5 7.15 v
er6 14.6 v
..er6 = 14.55 volts. You need to carry more digits.
er7 13.2 v
..er7 is zero volts as long as the switch is open.
er8 13.2 v
er9 2.81 v
er10 1.75 v
er11 4.56 v
r2 8.1 ohms
..R2 = 9.7/1.2 = 8.08333 ohms. Need more digits.
r6 10.1 ohms
..R6 is 14.55/1.43 = 10.1748 ohms. Need more digits.
r8 5 ohms
..R8 is 13.2/2.63 = 5.019 ohms. Need more digits again.
r9 2.56 ohms
..You got ir11 correct at 2.28 amps. Therefore the current through R9 is
..2.63-2.28 = .35 amps, not .912 amps. That makes the voltage across R10
..equal to 5*.35 = 1.75 volts, and then the voltage across R9 is
..4.56 - 1.75 = 2.81 volts.
..Next, the resistance of R9 is 2.81/.35 = 8.02857 ohms.

#### JoeJester

Joined Apr 26, 2005
4,077
You need to construct a table so you can be clear on KVL and KCL.

Keep your calculations consistent. If your going to two decimal places, keep it that way throughout.