Re: Newbie: Series/Parallel DC circuit current question

Each component in the network has some voltage across it.

I got it. I have to use the value of the voltage across that component
or section not what's left over from other parts. That's where I
went wrong. I have to use the 20v being used between junctions A & B
or that "section" of the circuit then use Ohms law to find the
value of current thru each branch.

To make sure I have it, let's redo this circuit without R5 & R6. So
if R5 & R6 didn't exist, we'd have just the 5ohm value of R1 then
the 10ohm value of the parallel section of R2, R3, R4 so then Rt would
be 15 ohms. And current would be 100v/15ohms = 6.666 amps. The voltage
drop of R1 = 6.666 amps * 5 ohms = 33.33 volts "consumed" by R1.
And the voltage drop of the parallel section of R2, R3 & R4 would be
6.666 amps * 10 ohms = 66.66 volts dropped or consumed by the parallel
section. So since 6.666 amps flows into the parallel section at
junction A, it must flow out at junction B. What is the current thru R2
you ask ? :) IR2 = 66.66volts / 30 ohms = 2.222. Multiply that by 3
since each branch is of equal resistive value and you have 6.666 amps
total.

Thanks I can stop banging my head against the wall...at least until I
come to the AC section !

j

.

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