Resistance in a series / parallel?

• dolpho
In summary, the equivalent resistance between points A and B is 1.56 Ohms. This is calculated by finding the parallel equivalent of R3, R4, and R5 and adding it to R6 in series, and then finding the parallel equivalent of that with R1 and R2. The final calculation is 1/2.2 + 1 + 1/8.97 = 1.56 Ohms.
dolpho

Homework Statement

Find the equivalent resistance between points A and B shown in the figure(Figure 1) . Consider R1= 2.2 , R2= 1.0 , R3 = 3.6 , R4= 5.0 , R5= 4.0 , and R6= 7.6 .

http://i.imgur.com/EBJmSvk.png

Homework Equations

For resistors in parallel, 1 / Req = 1/r1 + 1/r2 + 1/r3...etc
for resistors in series, Req= R1 + R2 + R3...etc

The Attempt at a Solution

So I think R3, R4 and R5 are parallel

1/3.6 + 1/5 + 1/4 = .72

and then I think we can combine R6, R1 and R2 into one since they are parallel also.

1 / 2.2 + 1/1 + 1/7.6 = 1.58

Now add up .72 and 1.58 since they are in series? = 2.3

Would appreciate any help, this is pretty confusing to me lol

dolpho said:
So I think R3, R4 and R5 are in a series.

Check this again.

Pranav-Arora said:
Check this again.

Ahhh I'm sorry I did a typo. I meant parallel

dolpho said:
Ahhh I'm sorry I did a typo. I meant parallel

Did you get the correct answer now?

Pranav-Arora said:
Did you get the correct answer now?

nope :( I'm not sure what I'm doing wrong

dolpho said:
So I think R3, R4 and R5 are parallel

1/3.6 + 1/5 + 1/4 = .72

0.72 is reciprocal of equivalent resistance of R3, R4 and R5. The equivalent of these three resistors is in series with R6.

Pranav-Arora said:
0.72 is reciprocal of equivalent resistance of R3, R4 and R5. The equivalent of these three resistors is in series with R6.

Ok I was just thinking that, so its actually.

.72 = 1 / Req

Req = 1.39?

Then we add that to R6 in a series? so

1.39 + 7.6 = 8.99

Now we add R6 to R2 and R1 which are parallel.

So... 1/2.2 + 1 + 1/8.99 = 1 / Req

= 1.24 = 1 / Req -----> 1 / 1.24 = .8?

dolpho said:
Ok I was just thinking that, so its actually.

.72 = 1 / Req

Req = 1.39?

Then we add that to R6 in a series? so

1.39 + 7.6 = 8.99

Now we add R6 to R2 and R1 which are parallel.

So... 1/2.2 + 1 + 1/8.99 = 1 / Req

= 1.24 = 1 / Req -----> 1 / 1.24 = .8?

I haven't checked out your calculations but your steps look fine to me.

Pranav-Arora said:
I haven't checked out your calculations but your steps look fine to me.

ARG...It's not .8...what the f...heck

dolpho said:
So... 1/2.2 + 1 + 1/8.99 = 1 / Req

= 1.24 = 1 / Req -----> 1 / 1.24 = .8?

Calculation mistake. 1/Req=1.56

Pranav-Arora said:
Calculation mistake. 1/Req=1.56

Finally! Got it right, thanks so much for your help!

dolpho said:
So... 1/2.2 + 1 + 1/8.99 =
= 1.24

Try that step again (btw, 8.97 is more accurate)

1. What is the difference between series and parallel resistance?

In a series circuit, the resistors are connected end to end, creating only one path for the current to flow. In a parallel circuit, the resistors are connected side by side, creating multiple paths for the current to flow.

2. How do I calculate total resistance in a series circuit?

In a series circuit, the total resistance is equal to the sum of all individual resistors. So, if there are three resistors with values of 10 ohms, 20 ohms, and 30 ohms, the total resistance would be 60 ohms.

3. How do I calculate total resistance in a parallel circuit?

In a parallel circuit, the total resistance is calculated using the formula: 1/Rt = 1/R1 + 1/R2 + 1/R3... where Rt is the total resistance and R1, R2, R3, etc. are the individual resistances. Once you have the sum of the reciprocals, take the reciprocal of that sum to get the total resistance.

4. What happens to total resistance in a series circuit as more resistors are added?

In a series circuit, as more resistors are added, the total resistance increases. This is because the current has to flow through each resistor, and each resistor acts as a barrier to the flow of current.

5. What happens to total resistance in a parallel circuit as more resistors are added?

In a parallel circuit, as more resistors are added, the total resistance decreases. This is because the current has more paths to flow through, and the more paths there are, the easier it is for the current to flow, resulting in a lower overall resistance.

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