# Question: Resistors connected in parallel/ series

1. Sep 20, 2009

### cathode

1. The problem statement, all variables and given/known data

1.The equivalent resistance of several resistors connected in parallel is always lower than the smallest valued resistor.

A) True
B) False

2. The following two circuits are equivalent at terminals A-B.

http://img6.imageshack.us/img6/1401/circuity.jpg [Broken]

A) True
B) False

3. For three resistors connected in series, the voltage across each resistor is equal even if the resistors have different values.

A) True
B) False

----

For number 1, I guessed True. I'm not sure but my reasoning behind is by KCL and KVL, voltages around a loop and current entering a node is zero at every instant. Again, I'm not sure about this one.

For number 2, I guessed False. The cicuit on the left, I used Ohm's law to find its voltage; 50(v). On the other hand, the right circuit doesn't have a resistor at all. So I thought since 50(v) is greater than 0 (v) {the right circuit}, the statement is false, which states two circuits are equivalent.

For number 3, I guessed True, because by the definition KVL, the sum of voltages around a loop equals zero.

Last edited by a moderator: May 4, 2017
2. Sep 21, 2009

### jambaugh

Note: I'm commenting about your reasoning but not saying any of your answers are wrong...or right. It's not sufficient to guess correctly but to know why which I assume you already appreciate.

Number 1 start with the smallest valued resistor by itself. If you add another in parallel will the total resistance go up or down?

Number 2 your reasoning looks good.

In number 3 there is no mention of a loop. You have three resistors in series. That means current flowing through any one flows through them all so the currents are equal. How does this relate to the voltages?

3. Sep 21, 2009

### cathode

For 1, the total resistance will go up.

For 2, I checked my textbook, and there is an example that shows a circuit and a equivalent circuit of its original circuit. Turns out, I was wrong. This one is actually True.

For 3, that means the total voltage passing through each resistor is equal to the total resistance of the three resistors.

R1 ---- R2 ---- R3
3ohms 4ohms 5ohms
3volts 4volts 5volts

Therefore, it's false!!

4. Sep 21, 2009

### jambaugh

Think of a stream analogue. If you dig a channel parallel to the stream do you get more or less flow? (more flow = less resistance, less flow = more resistance)?
I thought the question vague. Was there more detail about the meters?
Right!

Nothing beats a nice concrete example to clarify!

5. Sep 21, 2009

### CEL

About question number 2. You have an ideal current source. The current in terminal A will be 5A, whether there is a resistor in series or not.

6. Sep 21, 2009

### jambaugh

Ahhh! I didn't recognize that particular circuit element. That explains it. Does that make sense to you Cathode?

7. Sep 21, 2009

### mikelepore

Why don't you just try a few with a calculator? For example, the parallel combination of 10 ohms and 90 ohms would be 9 ohms. What general pattern emerges from such results?

8. Sep 21, 2009

### ramonegumpert

Hi Cathode,
for question

3. For three resistors connected in series, the voltage across each resistor is equal even if the resistors have different values.

A) True
B) False

>>> The answer should be B) False. Different resistor value gives different voltage as the current is the same in series.

9. Sep 21, 2009

### cathode

I'm still not understanding about question 1...

10. Sep 22, 2009

### CEL

For simplicity, consider two resistors: R1 and R2. What is the resistance of their parallel connection?

11. Sep 22, 2009

### cathode

is the relationship, the product over sum rule?

12. Sep 22, 2009

### mikelepore

Yes, if the number of resistors in parallel is exactly two.

If there are two, then the more general expression R = 1/[(1/R1)+(1/R2)] will also be equal to R1 R1 / R1 + R2. But don't try using the product over the sum if there are three or more resistors.

13. Sep 22, 2009

### CEL

Yes.
$$R_{eq}=\frac{R_1R_2}{R_1+R_2}$$
Suppose $$R_2$$ is the smaller resistor. Divide the numerator and the denominator by $$R_1$$. You get:

$$R_{eq}=\frac{R_2}{1 + R_2/R_1}$$.
Since $$R_2$$ is divided by a quantity greater than 1, the result must be smaller than $$R_2$$.
You can consider the parallel of three resistors as the parallel of one with the equivalent of the parallel of the other two and so on.

14. Sep 22, 2009

### cathode

ahh now I get it.
Thanks so much for the help, everyeon!