Understanding Simple Circuits: Answering Common Homework Questions

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The discussion focuses on understanding how changes in resistance affect voltage and current in simple circuits. Participants clarify that increasing resistance (R7) decreases the potential difference between points A and E, as well as the voltage drop across R4. They emphasize the importance of Ohm's law, which states that current is equal to voltage divided by total resistance, to analyze these relationships. Additionally, it is noted that decreasing resistance (R2) leads to an increase in current through R1. Overall, the conversation aims to simplify circuit analysis for homework questions.
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Homework Statement



http://img125.imageshack.us/img125/5434/asdfjr7.jpg

The Attempt at a Solution


Basically for this excercise I have to fill in the gaps with "increases", "decreases" or "stays the same".

Some of the parts I can do, others I am clueless in. It would be helpful if somebody could explain it to me in a very simple manner :smile:

a) If R7 increases, potential difference between points A and E increases. Assume no resistance in [PLAIN]http://img82.imageshack.us/img82/4954/26005668vm6.jpg and έ (the battery)

b) The same as in a) but with resistance between [PLAIN]http://img82.imageshack.us/img82/4954/26005668vm6.jpg and έ (the battery)
[I'm unsure what the answer is]

c) If R7 increases, voltage drop across R4 decreases

d) If R2 decreases, current through R1 _________ [unsure of answer]

There's a couple more, but I'll leave it at that for the time being so it's easier to "digest".
 
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For a and b, you need to write Ohm's law on A-->E (the lower part which has the battery), you will see how the voltage drop over AE depends on the current.

Hint: your answer on a is wrong

For other parts, you base on the variation of the current to justify the drop on each part.
 
Weimin said:
For a and b, you need to write Ohm's law on A-->E (the lower part which has the battery), you will see how the voltage drop over AE depends on the current.

Hint: your answer on a is wrong

For other parts, you base on the variation of the current to justify the drop on each part.

So in a) resistance increases, and we want to find V but how does current change?
 
Use ohms law to write an expression for the current -

current = voltage/total resistance. The voltage across any element is then easy to read off, V= current * resistance of element
 
Mentz114 said:
Use ohms law to write an expression for the current -

current = voltage/total resistance. The voltage across any element is then easy to read off, V= current * resistance of element

You probably think I'm dyslexic, but I'm still struggling...

V=I*R, no worries I understand that.
R7 is increasing and we want to find V, but what about current? Does the current remain the same? How are we to find V if I is not known?
 
Imagine a circuit comprising 3 resistors and a battery with zero resistance, which has voltage V. So, I = V/(r1+r2+r3). The voltage across r1 is given by V1= I*r1 = V*r1/(r1+r2+r3)

It's obvious now that if you increase r2 or r3, the voltage across r1 goes down.
 
Mentz114 said:
Imagine a circuit comprising 3 resistors and a battery with zero resistance, which has voltage V. So, I = V/(r1+r2+r3). The voltage across r1 is given by V1= I*r1 = V*r1/(r1+r2+r3)

It's obvious now that if you increase r2 or r3, the voltage across r1 goes down.

Ok yep, I understand that now.
So in relation to the example, r1=r7, r2=r5 and r3=r2?
 
I'd recommend writing out the equation for your circuit. You can combine the 2 parallel sets into one resistor which gives 4 in the current denominator.
 
Just thought I'd revisit this question.
After some thought, these are my rehashed answers

a) If R7 increases, potential difference between points A and E DECREASES. Assume no resistance in and έ (the battery)

b) The same as in a) but with resistance between and έ (the battery) DECREASES

c) If R7 increases, voltage drop across R4 DECREASES

d) If R2 decreases, current through R1 INCREASES
 
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