Question

I= (E)/(sqrt(x^(2)+R^(2)))
for x= 30*10^(3)
R= 40*10^(3)
E= 10

Answer 1

\[\large I=\frac{E}{\sqrt{x^2+R^2}}\]



\[\large I=\frac{10}{\sqrt{(30*10^3)^2+(40*10^3)^2}}\]



\[\large I=\frac{10}{\sqrt{900*10^6+1600*10^6}}\]



\[\large I=\frac{10}{\sqrt{2500*10^6}}\]



\[\large I=\frac{10}{\sqrt{2500}*\sqrt{10^6}}\]



\[\large I=\frac{10}{50*10^3}\]



\[\large I=\frac{10}{50*10^3}\]



\[\large I=\frac{1}{5*10^3}\]



\[\large I=\frac{1}{5000}\]


Note: alternatively, you can express the answer as \[\large I=2*10^{-4}\]

Answer 2

i have multiple choice anwsers and thats not one of them

Answer 3

what are your choices

Answer 4

probably best to post those along with the problem at the top

Answer 5

1) 2 mA 2)0.2mA 3) 0.5mA 4)0.0005A

Answer 6

oh sorry

Answer 7

not sure what the units are in the problem. Are those provided as well?

Answer 8

elctrical algebra

Answer 9

well 2 mA is 2 milliamps, which is 0.002 amps (note: milli means "thousandth")

0.2 mA is 0.2 milliamps, which is 0.0002 amps


So my guess is choice B since \[\frac{1}{5000}=0.0002\], but without the units, I can't be 100% certain.

Answer 10

reading you explinations i have to agree with you so thank you so much