Help with Circuit Problem!?

Question

Hi, I'll be taking physics final tomorrow and I'm just so confused on the circuit problems.

Could you tell me the convention of the relatioships among the V, I, and R?

I mean like..
1. the voltage drop across each resistor wheather it is in series or parallel are all same as the V of battery?

2. to find current in EACH resistor(parallel), I don't have to look for Req to calculate but to just use each resistance?

3. the sum of V drop across the series part of circuit and across the parallel part of circuit is same as total V?

4. How would you find the V drop across each resistor that are in series-parallel complex?

5. I just don't get when to find Req to calculate what
and when to separate series and parallel part of R to calculate what..

correct me if i'm getting all those wrong and please explain me some general conventions to follow in each possible case.
Thank you so much..I really want to do well on this test..:)

Answer 1

1. A voltage supply is assumed to be constant. For the problems you are solving, treat it as such. So whatever path you can trace from the positive terminal to the negative terminal will sum to the voltage of the supply. Always, never changes. If I put one resistor in parallel or a million, the voltage is always the same across each and every resistor. A resistor is a passive, linear device. That is, it has a functional relationship between voltage and current that is linear. The function is Ohm's Law: V=I*R
If I know any two of the variables, I can compute the third using this equation. I=V/R and R=V/I.

If the path you trace from the voltage supply has two resistors in parallel, and nothing else in parallel with those resistors, then there is a handy circuit reduction technique to compute the equivalent resistance.

I will derive that relationship here to help you understand the principles behind it.

Let's say there are 2 resistors R1 and R2 in parallel across a voltage V. They each have the same voltage, but different currents:
V=I1*R1
and V=I2*R2
Also, the equivalent resistance when calculated will also have the same voltage
V=I*Req where I is the sum of I1 +I2
or
V=(I1+I2)*Req
since V=I1*R1, then V/R1=I1
similarly, V/R2=I2
so
V=(V/R1+V/R2)*Req
divide out V, and divide both sides by Req
1/Req=1/R1+1/R2

Does that help?

2. I actually answered that above. Since the voltage is the same, then the current is just V/R for each resistor.

3. This is the same principle: Any path I can trace from the positive terminal to the negative terminal, the sum of the voltages in that path will equal the voltage of the supply.

Here's an analogy I hope will help. In Newtonian Physics you learned the concept of energy conservation. You studied the relationship between kinetic energy and potential energy. Think of voltage as potential energy. Think of a problem like:
Sally rode her bicycle on a roller coaster track that was frictionless. If she started at a height of h and got to the bottom her potential energy change was -m*g*h. If she climbed back up, or was shot out of a cannon back up, or rode a helicopter, whatever, when she got back to h, her potential energy change was m*g*h. Since she ended up where she started, the net change in potential energy was zero. Always, never changes in a closed system. Think of voltage the same way. No matter how I get from the positive terminal of the voltage source to the negative terminal of the voltage source, the net change in voltage is V, the supply voltage. When I go back up through the source to the positive terminal from the negative terminal, the voltage is -V. These two sum to zero. Always, no matter what path I took.

4. Once you understand voltage, the next step is to understand current. The analogy here is momentum. The starting momentum is always equal to the resultant momentum. Do you remember sum of m1v1= sum of m2v2?
Remember that velocity is a vector so you had to answer all kind of trick questions to make sure you got the directions of objects right in collisions?
Current is like that. At any node of the circuit the current entering the node is equal to the current leaving the node. Or, the sum of the currents at a node are equal to zero. Always. These two laws are the basis of circuit analysis. The rest is technique. Using these two laws, the resistor circuits you are analyzing can always be solved.

5. Did my explanation of how to calculate Req help?

I always start a circuit analysis by assessing what I know by inspection. A resistor in parallel with a voltage supply, I know the current to be V/R. If I see two or more resistors in parallel I just write down R1*R2/(R1+R2) and put a value in case I need it to sum voltages along a path.

The other element you may encounter in your test is a current source. These are tricky little devils because the current is constant, independent of the voltage across it. When I see one of these I always try to calculate the voltage across it if required to answer the question. Invariably it is required. Since the current is constant, any resistors in series with the current source will have V=I*R. So I can calculate voltages of the resistors and then hopefully sum them around some known voltage to calculate the voltage of the current source. These are always where the sign bookkeeping is critical.

Best of luck.

j