# Homework Help: Calculating Power Distribution in Circuit

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1. Aug 19, 2016

### bpetersen

1. The problem statement, all variables and given/known data
I am trying to calculate the power of Rload in the circuit attached (SEE ATTACHED CIRCUIT). The coupler I am trying to model is the Agilent 778D. Can someone verify that I calculate this out correctly?

2. Relevant equations
y dB = 10 * log10(x)
V = i * R
P = v * i

3. The attempt at a solution
i. Calculating the percent of power the coupler uses:
-0.6dB = 10 * log10( x )
-0.06 = log10( x )
10^( -0.06 ) = 10^( log10( x ) )
x = 87.10%
>> So the Coupler in the diagram consumes 12.90% of power seen at its input <<

Power seen at couplers input = Power_Source - Power_R1
So Power_Coupler = 0.1290 * ( Power_Source - Power_R1)

Power_Source = Power_R1 + Power_Coupler + Power_Load
Power_Source = Power_R1 + 0.1290 * ( Power_Source - Power_R1) + Power_Load
Power_Source = Power_R1 + 0.1290 * Power_Source + 0.1290 * Power_R1 + Power_Load
0.8710 * Power_Source = Power_R1 + 0.1290 * Power_R1 + Power_Load
0.8710 * Power_Sourse = 1.1290 * Power_R1 + Power_Load
0.8710 * V_Source = 1.1290 * V_R1 * i + V_Load * i
0.8710 * V_Source = 1.1290 * V_R1 + V_Load
0.8710 * V_Source = 1.1290 * i * R_1 + i * R_load
0.8710 * V_Source = i * ( 1.1290 * R_1 + R_Load )
V_Source = i * (1.1290 * R_1 + R_Load) / 0.8710

V_Load * i = V_Source * i * (R_Load * 0.8710) / (1.1290 * R1 + R_Load)

So if R1 = R_Load = 50Ohms

Thusly,

The power of the load = 40.91 % of the Power Source

y dB = 10 * log10 (0.4091) = -3.8817 dB

Conclusion:

The power distribution is as follows:
R1 = 46.19%
Coupler = 12.90%

and I should see a -3.8817 dB attenuation of the signal at the load

#### Attached Files:

• ###### Power Diagram.png
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2. Aug 20, 2016

### rude man

Did you include the power dissipated in the rf analyzer, which must also be a 50 ohm load?