Help with question

[Crackle]

The power amplifier in a transmitter has an efficiency of 60% and provides 50W of RF at its output. How much
power is lost as heat?
A 84W
B 30W
C 33W
D 20W

I cant get my head around this question, please can someone help explain.

The correct answer is given as 33W

Thanks
Mike

[Crackle]

I think I have got it, you need to do a sum with all the answers to see which is nearest 60%
So total power is
a) 50 / 134 x100 =37%
b) 50 / 80 x 100 = 62.5%
c) 50 / 83 x 100 = 60.2% ***
d) 50 / 70 x 100 = 71.4%

Is there an easier way rather than going through all the scenarios.

Thanks
Mike

[Big-Al]

Hi
  Yes an easy way IMHO to look at it is that 50 watts is the output ( this represents the 60% as Rf power, the remaining 40% must therefore be transformed into heat. therefore the power dissipated as heat is 40% )
  if 50watts equals 60% of input then 34 watts equals 40% of input. or 2/3rds of 50watts.
                           Hope you can follow my twisted logic....Regards Big Al

[Alistair D]

This way looks easier to me.

60% = 50 watts  

100% = (100/60)*50 = 83watts

 83 - 50 = 33watts

Al

[Crackle]

Hi Al and BIG Al
Yes I agree Al your method is easiest, I should have come up with that;
if 60% = 50w
1% = 50/60 = .83'w
so 40% = .83 x 40 = 33.3w

Thanks guys for knocking my brain into gear.

Mike

[Crackle]

The capacitance of a variable capacitance diode
A. increases with increasing voltage when forward biased.
B. decreases with increasing voltage when forward biased.
C. increases with increasing voltage when reverse biased.
D.* increases with decreasing voltage when reverse biased.

Answer is D. This one had my head spinning, till I looked it up.
Talk about making it more complicated than it need be.
Simple answer increases with less reverse bias.

Mike

[ppppenguin]

I looked at it and first thought of working out 100% like Alistair D, then I realised it was easier to use 40/60 or two thirds.

[Kalee20]

The capacitance of a variable capacitance diode
A. increases with increasing voltage when forward biased.
B. decreases with increasing voltage when forward biased.
C. increases with increasing voltage when reverse biased.
D.* increases with decreasing voltage when reverse biased.

Answer is D. This one had my head spinning, till I looked it up.
Talk about making it more complicated than it need be.
Simple answer increases with less reverse bias.

Strictly speaking, A is also correct! At zero bias you can measure a certain capacitance (as long as you are only applying a few tens of mV test voltage), if you forward bias with 100mV you'll measure more capacitance. When you get to 500mV the thing starts to turn on and capacitance becomes meaningless, though an instrument might read infinity.

[Crackle]

I passed despite my occasional confusion in the mocks.

There were no mind blowing challenges, I seem to remember this one being one of the most technical.
10 ohm and 20 ohm resistors in series, which one has the greatest voltage across it.

And where would you place a resistor to make the transistor turn on the light bulb. (diagram with base bias resistor missing.)

All pretty easy really.

Mike

[ppppenguin]

Congratulations. What comes next?