Finding current across resistors

  • #1
notmetalenough
11
0
I'm not quite sure how I'm supposed to be doing this problem.

Consider the circuit shown in the figure below.

R = 21.3 ohms

http://east.ilrn.com/graphing/bca/user/appletImage?dbid=411814203 [Broken]

Find the current in the 21.3 ohm resistor.


I combined (serial) R with the 5.85 ohm resistor, getting an equivalent resistance of 27.15 ohms. Then combined the middle (parallel) 11.7 and the other 5.85, getting an equivalent resistance of 3.9 ohms. Then I combined the 3.9 and the 27.15 as parallel resistors leaving me with a 3.41 ohm resistor and an 11.7 ohm resistor in series. which I added together to get a total resistance of 15.11 ohms.

I then divided the Voltage by the resistance to get the amount of current, and worked backwards applying it to each series part and then dividing it among the parallel components.

I ended up getting something along the lines of 0.283 A through the 21.3 resistor. Did I miss something somewhere? Did I combine these correctly?

Thanks
 
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Answers and Replies

  • #2
Chi Meson
Science Advisor
Homework Helper
1,874
10
The middle 11.7 ohm, the bottom 5.85 ohm, plus the 27.15 ohm series are all in parallel in this arrangement. An inverse sum of this parallel is 3.41 ohms.
 
  • #3
notmetalenough
11
0
Chi Meson said:
The middle 11.7 ohm, the bottom 5.85 ohm, plus the 27.15 ohm series are all in parallel in this arrangement. An inverse sum of this parallel is 3.41 ohms.

Yeah, I had that in there. The parallel set is then in series with the top 11.7, and then I still can't get the right answer.
 
  • #4
The 27.15 resistor is still in series with the 11.7 in the middle, not in parallel. It's better to combine the 3 in series so you have 3 parallel. The voltage over resistors in parallel are equal, so you can calculate the current over the equivalent resistor.
 
  • #5
Chi Meson
Science Advisor
Homework Helper
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Romperstomper said:
The 27.15 resistor is still in series with the 11.7 in the middle, not in parallel. It's better to combine the 3 in series so you have 3 parallel. The voltage over resistors in parallel are equal, so you can calculate the current over the equivalent resistor.

No it's not. The current splits at the junction near the "a" dot. It rejoins at the junction near the "b" dot.


The total resistance of the circuit is 15.11 ohms, so the total current is 1.33 amps.

The voltage drop through the top 11.7 ohm resistor is 15.56 volts, which leaves 4.54 volts across the parallel. 4.54 volts across the 27.15 ohm series says that 0.17 amps flows through this branch.
 

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