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aabbaa11

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- TL;DR Summary
- My question is what is the connection between equations on the power dissipated by resistor in DC and high frequency?

Hello all PF members!

I was wondering about the power dissipated by resistor at high frequency which is:

##P=Re\left \{ U\cdot I^{*} \right \}=Re\left \{ \frac{\left |V^{+} \right |^{^{2}}}{2\left |Z_{c} \right |}\cdot (1-\left |\Gamma \right |^{2} )\right \}##, (1)

where:

##Z_{c}## - characteristic impedance of the waveguide (transmission line) = ##\sqrt{\frac{R_{c}+ j\omega L_{c}}{G_{c}+ j\omega C_{c}}}##

##V^{+} ## - amplitude of incident wave (to the resistor)

##\Gamma## - reflection coefficient = ##\frac{Z_{L}-Z_{c}}{Z_{L}+Z_{c}}##, where ##Z_{L}## is load impedance.

These equations work fine at high frequency. The problem (and my question) is about the case of lower frequency and DC. When the frequency is going down the ##\left | Z_{c} \right |## is going to +inf, as well as ##1-\left |\Gamma \right |^{2} ## which is going to 0. Thus, the total active power is equal to 0 at DC. I can assume, that equation (1) represents the active power delivered only by wave, so for DC = 0. But as we know, the active power at DC is still transferred to the resistor:

##P=\frac{U^{2}}{R_{L}}## (2).

My question is: Which is the relation between (1) and (2)? Is equation (1) simplified somehow?

Thank you in advance!

I was wondering about the power dissipated by resistor at high frequency which is:

##P=Re\left \{ U\cdot I^{*} \right \}=Re\left \{ \frac{\left |V^{+} \right |^{^{2}}}{2\left |Z_{c} \right |}\cdot (1-\left |\Gamma \right |^{2} )\right \}##, (1)

where:

##Z_{c}## - characteristic impedance of the waveguide (transmission line) = ##\sqrt{\frac{R_{c}+ j\omega L_{c}}{G_{c}+ j\omega C_{c}}}##

##V^{+} ## - amplitude of incident wave (to the resistor)

##\Gamma## - reflection coefficient = ##\frac{Z_{L}-Z_{c}}{Z_{L}+Z_{c}}##, where ##Z_{L}## is load impedance.

These equations work fine at high frequency. The problem (and my question) is about the case of lower frequency and DC. When the frequency is going down the ##\left | Z_{c} \right |## is going to +inf, as well as ##1-\left |\Gamma \right |^{2} ## which is going to 0. Thus, the total active power is equal to 0 at DC. I can assume, that equation (1) represents the active power delivered only by wave, so for DC = 0. But as we know, the active power at DC is still transferred to the resistor:

##P=\frac{U^{2}}{R_{L}}## (2).

My question is: Which is the relation between (1) and (2)? Is equation (1) simplified somehow?

Thank you in advance!

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