# Complex Impedance of a voltage source and 2 resistors

1. Mar 7, 2016

### Potatochip911

1. The problem statement, all variables and given/known data
A real voltage source can be expressed as an ideal voltage source that is in series with a resistor that represents the inner resistance of the voltage source. This voltage source is a EMF and it is also in series with another resistor. Suppose both the EMF resistor $R_\varepsilon$ and the second resistor $R_2$ have complex impedance. Determine a) The current, b) the non-dc power that is dissipated in the second resistor and c) The requirements to achieve maximum dissipated power in the second resistor.

https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/80/80085-d91a852075e6af6d2347539788417783.jpg [Broken]
2. Relevant equations
$\tilde{v}=\tilde{i}z$
$z_ts=z_1+z_2+\cdots$
$z=|z|e^{j\phi}$
3. The attempt at a solution

For part a) the total impedance will be given by $z_t=z_1+z_2=(R_\varepsilon +ja)+(R_2+jb)=(R_\varepsilon+R_2)+j(a+b)$ so $|z|=\sqrt{(R_\varepsilon+R_2)^2+(a+b)^2}$ and $\phi=tan^{-1}(\frac{a+b}{R_\varepsilon+R_2})$ then from $\tilde{i}=\frac{\tilde{v}}{z}$

$$\tilde{i}=\frac{ve^{-j\phi}}{\sqrt{(R_\varepsilon+R_2)^2+(a+b)^2}} \\ i=\frac{v\cos(\phi)}{\sqrt{(R_\varepsilon+R_2)^2+(a+b)^2}}$$

now for b) would the non-dc power dissipated in the second resistor be $P=i\Im{(R_2)}=ibj$?

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Last edited by a moderator: May 7, 2017
2. Mar 8, 2016

### ehild

Only the real part of the impedance dissipates power. Check your last formula. Even the dimension is incorrect.

Last edited by a moderator: May 7, 2017
3. Mar 8, 2016

### Potatochip911

Whoops I also wrote $P=IR$ instead of $P_L=i^2R_L$, so it should be $P=i^2R_L$ then?

4. Mar 8, 2016

### ehild

Yes, if you mean that RL is the real part of z2. And you do not need the cosine term in the expression of the current as the voltage through RL is in phase with the current. You need to use the rms value of the voltage.

5. Mar 8, 2016

### Potatochip911

Yep & then I believe the power dissipated will be at a maximum when the two resistances are complex conjugates of the other since then the part of the denominator related to the imaginary numbers will go to 0.

6. Mar 8, 2016

### ehild

Yes, it is right.
Do not use the word resistance for a complex impedance. The impedance of the source must be complex conjugate of the loading impedance.