a) 100 ohm
b) 200 ohm
c) .5 meg
d) 5k ohm
e) 5m ohm
2007-12-13 09:05:53 · 13 answers · asked by Manh Cuong N 1 in Science & Mathematics ➔ Engineering
200 Ohms
2007-12-13 09:09:23 · answer #1 · answered by Jason 6 · 2⤊ 0⤋
the total resistance will be 5 m Ohm, since when resistances are connected in parallel, calculate the net reisitance as: (1/1000)*5 = (5/1000) = 5m Ohm.
2007-12-17 05:31:07 · answer #2 · answered by Priya S Pai 1 · 0⤊ 1⤋
b (200 ohm) series increases resistance 5x1000= 5k ohm
parallel decreases Resistance 1000/5=200 ohms.
2007-12-13 09:12:12 · answer #3 · answered by T C 6 · 0⤊ 0⤋
200 ohms! R = 1/(1/1000+1/1000+1/1000+1/1000+1/1000)
R = 1/(5/1000) = 1000/5 = 200.
2007-12-13 09:11:18 · answer #4 · answered by Know It All 2 · 0⤊ 0⤋
should it be .005 ohm.....cz the formula to find total resistance for a parallel series is 1/Re
1/Re = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5
= 1/1000 + 1/1000 + 1/1000 +1/1000 +1/1000
=5/1000
=0.005ohm
total resistance increase with the increase in resistors IN SERIES
whereas, decreases with the increase in resistor IN PARALLEL
2007-12-13 09:12:57 · answer #5 · answered by RaZoR 2 · 0⤊ 1⤋
b. 200 ohms 1/R = !/r1 +1/R2 +...+1/R5 = 1/1000 +...+1/1000
1/R = 5/1000 = 1/200 ---> R = 200 ohms
2007-12-13 09:10:01 · answer #6 · answered by nyphdinmd 7 · 0⤊ 0⤋
Since they are all the same, the answer is (1000 / 5) Ohms
In general if there were 5 different values you'd have to use this formula.
Rparallel = 1 / (1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5)
.
2007-12-13 09:09:12 · answer #7 · answered by tlbs101 7 · 1⤊ 0⤋
b) 200 ohm
2007-12-13 09:09:44 · answer #8 · answered by Anonymous · 0⤊ 0⤋
1/(sum(1/r) as r goes from 1 to n)
1/( 1/r1 + 1/r2 + 1/r3 + 1/r4 + 1/r5)
ok?
2007-12-13 09:09:57 · answer #9 · answered by cave_animal 2 · 0⤊ 0⤋
a) less than 1
2016-05-23 10:30:05 · answer #10 · answered by ? 3 · 0⤊ 0⤋