Things I cant work out

[lofty84]

When the emf in a conductor 0.15m long moving at right angles through a magnetic field, at a velocity of 2 m/s is 0.5 V, the magnetic flux density would be

I know the answer is 3.17 T but i dont know how to work it out


If a resistive and capacitive circuit has an impedance of 3 Ω and a resistance of 2.5 Ω, it's power factor would be

if anyone knows why the answers 0.83 leading could you show me why

If a resistive and inductive circuit has an impedance of 1 Ω and a resistance of 0.6 Ω, it's power factor would be

i know the answer is 0.6 lagging but i dont know why

please if anyone can show me the maths it would be appreciated

 

[lofty84]

This site use to be helpfull, whats happened

 

[SirKit Breaker]

This site use to be helpfull, whats happened

Would love to help you mate, but to be honest its years since i did all that stuff. To be honest now i haven't got a clue. Try same post in Testing and inspection forum, there are some clever lads in there that teach etc and will be a bit more clued up with these type of calcs who may be able to help you. Failing that try the IET forums.

Cheers........Howard

 

[ringer]

I can tell you that the maths is resistance divided by impedance, but cannot tell you WHY you do it that way.

 

[ezzzekiel]

its taken from the triangle same as power triangle where you have resistance reactance and impedance and the difference of the angle results in the pf, think its the opposite over the hypot so 2.5/3 = 0.83pf

would need to draw it out but works same as tru power/ apparent power etc.

got the three triangles somewhere see if i can dig them out

This site use to be helpfull, whats happened

maybe its time you found somewhere else to contribute

 

[johnboy6083]

i can answer the last 2 questions for you, the equation is power factor (also called cos Φ) = R/Z

R is resistance and Z is impedance

 

[ezzzekiel]

using the resistance triangle:
impedance is the hyp
resistence is the opposite
and reactance (tangent?)

so opp / hyp = 2.5/3 = 0.83

pf is the ratio between real power and apparent power and is found by the phase angle using power, voltage or resistance triangles (i thinks)

 

[johnboy6083]

as for yuor first questuion, i have dug deep in my old course notes and found that the equation you require is e= Blv

e = induced emf (volts)
B= flux density (tesla)
l= length of conductor(m)
v=velocity(metres per second)

you need to transpose it, but i using your answer, i cant get it to fit. I was never any good at transposing anyway.

 

[Kirchoff1]

First Question e=Blv*sin(angle) as it is 90 degrees (right angle) becomes e=Blv
rearranging becomes B= e / lv

 

[Phil]

This site use to be helpfull, whats happened

the site is still helpful, not everyone is on on a weekend. have a browse through some of ambers posts for aprentises as there is loads on there for the calcs

 

[johnboy6083]

First Question e=Blv*sin(angle) as it is 90 degrees (right angle) becomes e=Blv
rearranging becomes B= e / lv

i transposed the formula to that, but it didnt fit with his values.

maybe we both got the transpostion wrong, or the values are wrong.

john

 

[lofty84]

thanks to everyone who has helped me to work these out it is very much appreciated