Circuit Problem

[Soccerdoc11]

R1 = R2 = R3 = R6 = 10 Ω, R4 = R5 = 20 Ω. Unless otherwise indicated, the battery can be assumed to be ideal (no internal resistance).

I'm wondering if someone can show me possibly an easier way on how to find the total resistance when all switches are closed.. We never did circuits in my physics class so sometimes these mess me up.. I'm hoping someone else looks at it a different way than the Kaplan explanation.. Thanks in advance.. The answer is 20 ohms.

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[HawkeyePostOp]

There's a beautiful logic to it that's kind of annoying to explain. I've done electrical/electronic engineering and it's still not totally intuitive, but kinda like a sudoku. Basically you start by simplifying parts in a small section of the larger scheme and then you build everything into a giant expression for one resistor - so the resistor you are considering keeps expanding until you have the circuit.

First, add numbers 3 and 6 - they don't have nodes nor do they split current in between them(they have the same current, so add and get 20 ohms). The sum of 3 and 6 is parallel with 5.
so you got (20||20) as an expression

notice R2 has to carry the same unsplit total current that the (20||20) resistor is carrying, so add it.
the expression is now ((20||20) + 10)

notice this system is in parallel with the R4.
((20||20) + 10) || 20

and finally, R1 has to carry the same current as this new system, so add it
[((20||20) + 10) || 20] + 10

now simplify

you could use 1/Rtot = 1/Ra + 1/Rb + 1/Rc.....etc for parallel, but since you have only two resistors in parallel at a time, it's easier to use the shortcut

Rtot = (Ra*Rb)/(Ra+Rb)

[((20||20) + 10) || 20] + 10 works out to 20 ohms

 

[Soccerdoc11]

Thanks that helps - my initial intuition was that all 6 were in pairs of resistors in series which were then in parallel with the others.