Complex Impedance of a voltage source and 2 resistors

  • #1

Homework Statement


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]

Homework Equations


##\tilde{v}=\tilde{i}z##
##z_ts=z_1+z_2+\cdots##
##z=|z|e^{j\phi}##

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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Answers and Replies

  • #2
ehild
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Homework Statement


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]

Homework Equations


##\tilde{v}=\tilde{i}z##
##z_ts=z_1+z_2+\cdots##
##z=|z|e^{j\phi}##

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##?
Only the real part of the impedance dissipates power. Check your last formula. Even the dimension is incorrect.
 
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  • #3
Only the real part of the impedance dissipates power. Check your last formula. Even the dimension is incorrect.
Whoops I also wrote ##P=IR## instead of ##P_L=i^2R_L##, so it should be ##P=i^2R_L## then?
 
  • #4
ehild
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Whoops I also wrote ##P=IR## instead of ##P_L=i^2R_L##, so it should be ##P=i^2R_L## then?
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.
 
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  • #5
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.
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
ehild
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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.
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.
 
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