Recent content by charger9198

Thanks rude man, I now can confirm I get the same as you L=1.872 Hy C=2.165 nF

Thanks, i agree i did think that. do you mind me asking what equation you transposed for L? All I can come up with is that involving capacitance

A series RLC circuit is connected to a 5 V supply, the frequency of the supply is adjusted to give a maximum current of 11.9 mA at 2.5 kHz. The Q factor is 70. Determine the component values of the circuit. R= 5/.0119=420.2 Ohm Q = (1/R)*(sqrt L/C) 70 = (1/420.2)*(sqrt L/C) 70/(1/420.2)=...

Wow thanks for that Ryan, I find it fascinating

I was watching a programme on anti matter 'discovery' whereby a guy claimed that research is going on to create this... The way this guy described the energy was that if a 'anti matter' clone of himself was to touch him, both him and the surroundings would be obliterated.. He claimed that...

Thanks for you help as always :)

Think I've got it; i=4.34(cos(0.76699)-jsin(0.76699) i = 4.34(0.720003-0.693971) i = (3.12481-3.01183j) So Z=120/(3.12481-3.01183j) Z=19.9079 + 19.1882j Minus ten from real gives 9.9079 ohms This better? Are my answers ok for the other parts of the question?

Ok thanks gneill so ill use ϕ=acos(power factor) =acos (.72) = 0.76699 then; i = 4.34(cos(0.76699)-jsin(0.76699)) i = 4.339566 Zt=120/4.339566 =27.28723 (minus 10) =17.28723

For the circuit given in the power factor is 0.72 lagging and the power dissipated is 375 W. Determine the: (1) apparent power (2) reactive power (3) the magnitude of the current flowing in the circuit (4) the value of the impedance Z and state whether circuit is inductive or...

Thanks for your help

so p=(I^2)R thus (0.8409^2)*1 = 0.70711 W (-0.4545^2)*2 = 0.41314 W (0.39^2)*3 = 0.4563 W Hopefully that's it!

Of course... R1 = 0.8409^2 / 1 = 0.7071 W R2 = -0.4545^2 / 2 = 0.2066 W R3 = 0.39^2 / 3 = 0.4563 W Am i on the right lines with this?

Calculate the power dissipated in R1, R2 and R [/b] Previously i worked out the Current I1, I2 and I3 so i planned to use this to calculate the power dissipated; For current i got; I1= 0.8409 A I2= -0.4545 A I3= 0.39 A I used I^2/R to get the following answers R1 =...

I think your I2 should be -0.4545 which changes your I1 to to 0.8409 giving I3 = 0.39 A. This would affect your answer when working out the power dissipated on each resistor wouldn't it?

Thanks for both your help guys; technician: I like your method, thanks for the input:)