What is the current of the coil while it is operating?
3.3 x 10-2 A
3.3 x 10-1 A
3.0 x 10-1 A
3.0 x 101 A
2007-04-20 07:13:01 · 9 answers · asked by Sesily E 1 in Science & Mathematics ➔ Physics
Current I = Voltage V / Resistance R ( I = V / R)
V = 120 and R = 4 so, I = 120 / 4 = 30 amps which is same as 3 X 10^1 A
2007-04-20 07:17:20 · answer #1 · answered by Swamy 7 · 1⤊ 0⤋
v = i x r
i = v / r = 120 V / 4.0 ohms = 30 V/Ohms = 30 amps
= 3 x 10^1 A
2007-04-20 07:36:59 · answer #2 · answered by Dr W 7 · 0⤊ 0⤋
Current = 120/4=30 Amperes but actual current might be smaller than this figure due to increase in resistance of heater element with the temperature.
2007-04-20 07:29:33 · answer #3 · answered by dwarf 3 · 0⤊ 1⤋
120v divided by 4 = 30 amps or 3.0 x 10^1
V = I x R or V/R = I
2007-04-20 07:17:25 · answer #4 · answered by Mikee 3 · 0⤊ 0⤋
I=V/R
I=120/4=30 =3.0 x10^2 A
2007-04-20 07:16:42 · answer #5 · answered by Edward 7 · 0⤊ 5⤋
it is quite easy,
V=I.R
so ; 120 = 4.R
R = 3.0 X 10^1 amper
2007-04-20 07:31:41 · answer #6 · answered by Hurricane 5 · 0⤊ 0⤋
warmth absorbed=mxc8deltaT m=21 c=a relentless i forgot the cost for water (i'm 29 !! ) delta T is temperature improve warmth absorbed=100percentheat released(by resistor warmth released=I^2 x R x time =fifty two.5*fifty two.5*2*3 hundred=answer from element b? so 1653750/mc= delta T and c for water is 4190J/kg*ok
2016-11-26 00:45:01 · answer #7 · answered by ? 4 · 0⤊ 0⤋
V = IR
Voltage = Current * Resistance
So I=V/R
You do the math.
2007-04-20 07:15:37 · answer #8 · answered by Anonymous · 1⤊ 0⤋
c
2007-04-20 07:18:37 · answer #9 · answered by technicanb 4 · 0⤊ 2⤋