Show how you should connect nine 100Ω resistors each rated at 10 watts so that the combination will still be equivalent to 100Ω but with a much higher power rating of 90 watts.
2007-08-30 02:05:54 · 5 answers · asked by andrew 1 in Science & Mathematics ➔ Engineering
3 branches in parallel of 3 in series. Each of the 3 in series is 300 ohms, and 3 of those in parallel is 100 ohms.
2007-08-30 02:32:30 · answer #1 · answered by Gary H 6 · 1⤊ 0⤋
I can tell it but I am not sure how to draw it on Yahoo. You want to make 3 meshes of resistors. Sketch them on a sheet of paper after you read this.
Call the resistors R1,R2,R3,R4,R5,R6,R7,R8,R9.
Say that each one has two ends, called A and B.
Connect R1, R2, and R3 in series with each other. You have a first set of 3.
Connect R4, R5, and R6 in series with each other. You have a second set of 3.
Connect R7, R8, and R9 in series with each other. You have a third set of three.
Now connect them by putting R1B and R4A together and R3B and R6A together. You now have two connected loops.
Now connect the last group by putting R4B and R7A together and R6B and R9a together. Now there are three connected loops.
The open end is where you should see 100 Ohms. Each resistor should carry 10 watts for the total of 90 watts.
2007-08-30 03:45:42 · answer #2 · answered by Rich Z 7 · 0⤊ 0⤋
By connecting the nine resistors as three parallel paths of three series resistors each. The net equivalent resistance is still 100 ohm, but now, since each resistor can dissipate 10W, the entire arrangement can dissipate 90W.
2007-08-30 02:36:37 · answer #3 · answered by dansinger61 6 · 1⤊ 0⤋
CONNECT THE RESISTORS IN PARALLEL (ALL ENDS TOGETHER ON ONE END ALL OTHER ENDS TOGETHER ON OPPISITE END RESISTANCE WILL REMAIN AT 100 OHMS WATTS WILL MULTIPLY TIMES 9 OR 90 WATTS
2007-08-30 02:53:18 · answer #4 · answered by Anonymous · 0⤊ 2⤋
try to connect it and test with multimeter
2007-08-30 02:41:37 · answer #5 · answered by Rofaif 1 · 0⤊ 2⤋