General > General Technical Chat
The Art of Electronics: Solutions to Exercises?
timofonic:
I still don't own the book, but it's a very interesting idea! I mostly learn a lot better by looking at solved and very well explained exercises than reading theory.
What about a guideline to solutions for each exercise? I propose the following:
- Step by step.
- Detailed but easy to understand explanations, with references to book or any other requirement.
- Provide SPICE simulation files if possible and can help in some way.
- Write formulas in MathML or LaTeX, like in allabourltcircuits.
- Preference to use SVG over raster formats.
Mati256:
Looks like the solutions are nowhere to be found so I want to post here and see if anyone can help me.
I'm currently working on Excersise 1.10:
Given a voltage divider where Vin = 30V, R1 = 10K and R2 = 10K find:
a) Output voltage.
b) Output voltage with a 10K load.
c) Thevenin circuit.
d) Same as b but with Thevenin circuit.
e) Power dissipated in each resistor.
My answers:
a) Vout = 15V (Vin x (R2/(R1+R2)))
b) Vout = 10V (Same as a but R2 in parallel with Rload = 5k)
c) Vth = 15V Rth = 5k
d) Vout = 22mV (Vth x (Rth/(Rth+Rload)))
e) I'm not sure about this one. With load it should be .01W and without load 22mW?
I hope you can give me a hand here. Thanks!
akos_nemeth:
--- Quote from: Mati256 on February 13, 2017, 11:49:18 pm ---Looks like the solutions are nowhere to be found so I want to post here and see if anyone can help me.
I'm currently working on Excersise 1.10:
Given a voltage divider where Vin = 30V, R1 = 10K and R2 = 10K find:
a) Output voltage.
b) Output voltage with a 10K load.
c) Thevenin circuit.
d) Same as b but with Thevenin circuit.
e) Power dissipated in each resistor.
My answers:
a) Vout = 15V (Vin x (R2/(R1+R2)))
b) Vout = 10V (Same as a but R2 in parallel with Rload = 5k)
c) Vth = 15V Rth = 5k
d) Vout = 22mV (Vth x (Rth/(Rth+Rload)))
e) I'm not sure about this one. With load it should be .01W and without load 22mW?
I hope you can give me a hand here. Thanks!
--- End quote ---
Hi Mati256,
My results:
a) 15V
b) it will be a 10K:5K divider with the load, so Vth= 10V
c) Vth=15V, Rth=5K
d) this should be the same as b)
c) without load:
I=30V/20KOhm=0.0015A
P=I^2*R=0.0225W=22.5mW
with load:
I = 15V / 15KOhm=1mA (this will be at the voltage source)
P1 = I^2*R = 10mW (the upper 10K resistor)
P2,3 = (I/2)^2*10K=2.5mW (the current is split equally between the two lower 10K resistor)
Regards,
Ákos
Mati256:
Thanks for your answer. I still don't get part d. :-//
Can anyone elaborate?
EDIT: Never mind, I was using the wrong formula. Correct formula should be Vth x (Rload/(Rload+Rth))
akos_nemeth:
--- Quote from: Mati256 on February 14, 2017, 10:34:12 pm ---Thanks for your answer. I still don't get part d. :-//
Can anyone elaborate?
--- End quote ---
Hi Mati256,
Maybe the image helps. The original circuit is marked with "A". We can use the Thévenine equation to find the equivalent circuit "B". Then we add the load resisitor. The voltage at the loading point (marked with red) can be calculated with the simple voltage divider equation. Circuit "C" is the original circuit with the load resisitor and shows that the calculations were correct (same voltage and current ).
Regards,
Ákos
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