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Power Dissipation Problems

A signal generator outputs a sine wave signal with a frequency of 1000 Hz and an rms voltage of 20.0 V. The output of the spectrum analyzer is connected to a 200 ohm resistor.

a. What is the power dissipated in the 200 ohm resistor in watts, in dBW, and in dBm?

b. An amplifier with a gain of 15 dB over the frequency range 0 Hz to 10 kHz is placed in the circuit between the signal generator and the resistor. What is the power dissipated in the 200 ohm resistor in watts and dBW now?

c. An attenuator with a fixed attenuation of 6.0 dB over the frequency range 0 Hz to 10 kHz is placed in the circuit between the amplifier and the resistor. What is the power dissipated in the 200 ohm resistor in watts and dBW now?

d. The attenuator in part (c) above is replaced by a low pass filter. The filter has an attenuation of 3.0 dB at 500 Hz and an attenuation above 500 Hz of 12 dB per octave. What is the power dissipated in the 200 ohm resistor in dBm now?

Solution Preview

a. Power in watts is:
P = V^2/R
P = (20)^2/200
P = 2W

In dBW:
P(dBW) = 10 log 2
P(dBW) = 3.01 dBW

In dBm:
P(dBm) = 10 log (2/1mW)
P(dBm) = ...

Solution Summary

This posting contains the solution to the given problems in simple equation format. Each answer lays out the power dissipation equation and shows how to calculate the answer.

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