[FONT="]The electronics teacher is regurgitating the same questions, year after year.
It certainly would not take 2 hours to answer the 5 questions.
Very few transistors have a collector-emitter voltage of 0.6v.
Let me know of a LED that drops 1.4v.
Do you realise meter "2" is connected the wrong way.
Do you realise the 12v battery symbol is incorrectly drawn.
I don't know that the "+" and "-" on the zener diode is supposed to represent.
For Q2: The first thing you do is move the waveform up or down the screen so that the bottom of the waveform is sitting on one of the screen lines (graticules). This makes it easier to read the size of the waveform.
Secondly, I would write: the resistor and capacitor should be placed in a vertical line so that they represent what is commonly known as a "time-delay circuit." - with the "pick-off" point in the middle.
What is the reading of meter 3: If it is a high impedance voltmeter, the reading is 10v.
If is a low-impedance meter, the reading will be less than 10v.
If it is an ammeter, the reading will be 0.6mA
If is it a microammmeter, the needle will hit the stop.
If it is an ohm-meter, the needle will hit the "stop."
As you can see, the answers to these questions can be much more complex than you think. Normally, you assume a "current-reading-meter" has no voltage-drop across it and a voltmeter draws no current from the circuit.
[/FONT][FONT="]How much power is dissipated in the zener diode will depend on the impedance (resistance) of the meters in the circuit. [/FONT][FONT="]If you neglect the impact of the meters, you will have to say the current though the 6v zener is the same as the current flowing through the 1k resistor, which is 6mA. Thus the power [/FONT][FONT="]dissipated in the zener diode will be 0.06mA x 6v = 0.36w or 36mW[/FONT]
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