Find current at high/low frequency

  • Thread starter Thread starter Acuben
  • Start date Start date
  • Tags Tags
    Current Frequency
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 4K views
Acuben
Messages
61
Reaction score
0

Homework Statement


find the rms current delievered by the 45v (rms) power supply when
a) the frequency is very large
and b) the frequency is very small.

answer: a) 225mA, b) 450mA

http://yfrog.com/jucurrentqg

http://img714.imageshack.us/img714/1592/currentq.gif

Uploaded with ImageShack.us

Homework Equations


w= angular velocity (supposed to be omega)
L= inductor
C= capacitor
j= complex coefficient = sqrt(-1)
Zl= resistance of inductor
Zc= resistance of capacitor
Ztot= total resistance
R= resistance of resistor
P= Power

Xl=wL
Xc=1/wc
Zl=j*Xl
Zc=-j*Xc
Ztot= R + Zl + Zc = R + j(Xl-Xc)



The Attempt at a Solution



with frequency just being high, how am I supposed to get these exact numbers without letters/symbol for I? o.o I tried the calculation and omega did not cancel out either.
 
Last edited by a moderator:
Physics news on Phys.org
The key idea is that in the case of very high or very low frequency, you can take the impedance (resistance) of the inductor and capacitor to be ~infinite, respectively. Does it make sense why that's true (as an approximation at least)?

In effect, this eliminates one of the two loops in the circuit for each case.
 
zhermes said:
The key idea is that in the case of very high or very low frequency, you can take the impedance (resistance) of the inductor and capacitor to be ~infinite, respectively. Does it make sense why that's true (as an approximation at least)?

In effect, this eliminates one of the two loops in the circuit for each case.

ah yes it does, at first I thought the resistance would skyrocket to infinity / 0
thank you
 
Last edited: