Complex current calculation in Parallel R L Circuit

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
35 replies · 7K views
I think I will give up for now .. I looked at it too much that I cannot come up with anything that makes sense ... Thanks a lot for all the effort you made to explain me :)
my last shot is when jwC is equal to jEwC so I(t) = jEwC + E/(R+jwL) or
(R^2+(wL)^2)%20).gif
 
Physics news on Phys.org
gl0ck said:
my last shot is when jwC is equal to jEwC so I(t) = jEwC + E/(R+jwL)

This is right.. keep going when you're ready.

I = E (jwC + 1/(R+jwL)) = E ((jwC)(R+jwL)+1)/(R+jwL)

(I found a common denominator and added)

Multiply that out and see if you can find a C that will minimize |I|. The j*j you see in the numerator above will introduce a negative quantity that will let you make |I| smaller.
 
Thanks a lot man but I have to go to sleep.. I hope tomorrow to find the right solution .. For now I give up..
Thanks for the efforts both of you
 
gl0ck said:
I think I will give up for now .. I looked at it too much that I cannot come up with anything that makes sense ... Thanks a lot for all the effort you made to explain me :)
my last shot is when jwC is equal to jEwC so I(t) = jEwC + E/(R+jwL) or
(R^2+(wL)^2)%20).gif

Good. Separate the total real and imaginary components and factor out the common bits:
$$I = E \left[ \frac{R}{R^2 + (ω L)^2} + j ω \left(C - \frac{L}{R^2 + (ω L)^2}\right)\right]$$
What value of C will minimize the magnitude of the whole?
 
Thanks! So when I plugged in the numbers It gets really nasty. First can I use w = 2Pi and to treat jEwC as j220*2*Pi*C? also w*L to be 12.57 because jwL = j12.57 Ohms
so to the equation I = 220 [ R/(R^2+12.57^2) + j2*Pi(C - 0.4/(R^2+12.57^2)]
I= 220[8/(64+158.0049) + j220*6.2830 ( C - 0.4/(64+158.0049)]
I = 7.92+(220jwC - 220*0.0018jw)
I= 7.92 + j1382.26C - j2.4807C
I=7.92+j1379.78C

I hope to do the right calculations

Thanks
 
The impedances of reactive components (inductors, capacitors) change with frequency. Furthermore, they change in opposite directions so that inductor impedance goes up with increasing frequency while capacitor impedance goes down. So no, in general you can't just set ω to a convenient value like ##2\pi##; you must use the value specified.

For part (a) you just need to derive the expression for the current, you don't have to plug in any numbers.

For part (b) you need to first determine an expression for C that will minimize the magnitude of the current. Use the given hint.