Engineering Complex current calculation in Parallel R L Circuit

[gl0ck]

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

 

[aralbrec]

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.

 

[gl0ck]

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

 

[gneill]

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

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?

 

[gl0ck]

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

 

[gneill]

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.