Calculating R, L, and Optimal C for e(t) and i(t) in a Power Circuit

[MissP.25_5]

I have done no. 1, but I don't how to do no. 2. Someone, please help.

Power source voltage e(t) = sqrt(2)*20sin2t [V],
current i(t) = 5sin(2t - pi/4) [A].

1. Find the values of resistance R, and inductance L.
2. Let V be the phasor of voltage v(t). Find the capacitance value of the capacitor C that maximizes the magnitude of V.

 

[rude man]

You know R and L so you can come up with the voltage V1 between L and R.

You can also write an expression for the voltage V2 between the 25 ohm resistor and C. This will of course be a function of E and C.

Then maximize (V2 - V1) with respect to C. (The value of C will not be a function of E).
All V are phasors. E is the phasor of e(t).

 

[MissP.25_5]

You know R and L so you can come up with the voltage V1 between L and R.

You can also write an expression for the voltage V2 between the 25 ohm resistor and C. This will of course be a function of E and C.

Then maximize (V2 - V1) with respect to C. (The value of C will not be a function of E).
All V are phasors. E is the phasor of e(t).

How do I write an expression for the voltage V2 between the 25 ohm resistor and C? I don't know the current across it.

 

[rude man]

How do I write an expression for the voltage V2 between the 25 ohm resistor and C? I don't know the current across it.

The current phasor through (not "across") C and the 25 ohm is E/Z where Z = 25 -j/ωC.

 

[MissP.25_5]

The current phasor through (not "across") C and the 25 ohm is E/Z where Z = 25 -j/ωC.

So now I get V1-V2. Next, do I write the function of d(v1-v2)/dC = 0?

 

[MissP.25_5]

The current phasor through (not "across") C and the 25 ohm is E/Z where Z = 25 -j/ωC.

Can you check this, please? Is this correct?

 

[rude man]

So now I get V1-V2. Next, do I write the function of d(v1-v2)/dC = 0?

Absolutely right! You might want to make sure it's a maximum, not a minimum or inflection point).

 

[rude man]

Can you check this, please? Is this correct?

Sorry, that would go a bit beyond what we're supposed to do.