Understanding grounding - very basic question

Discussion in 'Homework Help' started by phenohm, Nov 12, 2006.

Nov 12, 2006
10
0
Hello! I'm self-taught, not taking a class, and I am just starting out. I was hoping someone could give me a hand with these two questions I don't understand from the Socratic Electronics Project. Someone let me know if my picture upload doesn't work...

q1 - Is V(out) not electrically common with the ground? How do you measure the voltage drop without a loop circuit? V(total)-V(pot)? Is the added resistor there in parallel with V(out)? How do you calculate a total unknown (V(out)) in parallel with a known resistor?
q2 - In question 7, why is the ground connection in the loop on the right just after the source, pre-load? How will that circuit function? And, if it can't function, i.e. if it's shorted, how are we measuring voltage across it?

Thanks a lot!

File size:
45.7 KB
Views:
42
File size:
49.2 KB
Views:
26
2. hgmjr Moderator

Jan 28, 2005
9,030
214
Greetings phenohm,

As regards the problem concerning the potentiometer, it is a straightforward voltage divider with the some parallel resistance calculation thrown in. You can simply replace the potentiometer with two resistance whose value depends on the position of the wiper.

Another way to tackle the problem would involve calculating the Thevenin's equivalent resistance and voltage at the wiper of the pot without the external resistor attached. If you want to take this route, I suggest you review the material in the AAC Tutorial section on Thevenin's Method.

Since you have taken it upon yourself to learn Electronic Theory on your own, you may want to take advantage of the full range of tutorial material that is available at www.allaboutcitcuits.com.

hgmjr

3. kubeek AAC Fanatic!

Sep 20, 2005
4,689
806
In q7, the position of the ground in the circuit is irrelevant, it is just a common point for measuring voltage. This means that by moving the ground to any point of the circuit (still talking about one halve of the of q7 circuit separately) won´t change anything, the circuit will also function with no ground at all.

the second part is just substracting the two results, like in no.6. Notice that voltage on point A is positive from ground and B is negative from ground, bacause in B part the ground is connected to the + pole of the battery.