Okay. I got two different answers for these and I don't know which one is right. If you will please help by letting me know which answer is right and why, I'd very much appreciate it.
This is the figure:
http://www.webassign.net/pse/p28-09alt.gif
where R = 20ohm
1. Find the current in R (20ohm resistor)
2. Find the potential difference between points a and b
Answers I got:
One says that the first 3 resistors on the top are all in parallel, and then the total of that is in series with the other two that have 20 and 5 ohm, so the total is 27.5
The other answer is that in the resistor at the very top, the current goes through the wire then meets a junction and seperates into say, i1, i2, i3 and then they all meet again at the junction near point a. So the bottom 2 resistors are in series, but then the total of that and the two resistors above it are all in parallel, then in series with the very top one.
So we have 1/25+1/10+1/5
then r= 2.94
added to 10
so 12.94
I am extremely puzzled. Please explain to me what is the right answer and why its correct.
Thanks.
Answers & Comments
Verified answer
"first 3 resistors on the top are all in parallel"
That is NOT true, because one has a battery in series.
I'm assuming the + junctions to the left and right of the middle 10 ohm resistor are junctions not crossovers. They should be drawn with dots such as a and b are drawn.
Your second answer looks correct.
You may be better off redrawing it. If you can't see that the 5 and 20 are in series, you need to go back to basics.
5 and 20 are in series, total 25.
That is in parallel with 10 and 5 for a total of
1/R = 1/25 + 1/10 + 1/5 = 0.1+ 0.2 + 0.04 = 0.34
R = 2.94 ohms.
That is in series with the upper 10, for a total R of 12.94 ohms
Total current is I = E/R = 25/12.94 = 1.932 amps
That produces a voltage across the 3 in parallel of
E = IR = 1.93 x 2.94 = 5.68 volts
And that produces a current in the 25 ohms of
I = E/R = 5.68 / 25 = 0.227 amps, which is your answer.
You've got shorted out the circuit. The bulbs have a resistance, the wire does no longer (well, it does, but it's much lower) The electrical power takes the path of least resistance, and the lights go out.
The second one is correct.