Find the new resistance if its length and D are doubled

In summary, the mistake in the attempt at a solution is that A2 should be equal to A1 multiplied by 4, not divided by 4. Additionally, the correct equation for R2 is R1/2. This is because when only the diameter is doubled, the area increases by a factor of 4, so it's like having 4 original resistors in parallel. When only the length is doubled, it's like having 2 original resistors in series. When both are doubled, it's like having 8 original resistors in parallel and 2 in series, resulting in R2 being equal to half of R1.
  • #1
Fatima Hasan
319
14

Homework Statement


Capture.png


Homework Equations


ρl = RA
A = π r2

The Attempt at a Solution


l = RA (ρ is constant)
A1 = π (D/2)2
= (πD2)/4
A2 =π (2D/2)2
= πD2
= A1 /4
l2 = 2*l1
##R=\frac{l}{A}##
R2 = (2l1)/(A1/4)
= 8 l /A
= 8R1

Could someone tell me where is my mistake ?
 

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  • #2
Fatima Hasan said:
Could someone tell me where is my mistake ?

Trying to do this in one bite.

What happens if only its diameter is doubled? (Hint: the area goes up by 4, so it's like having 4 original resistors in parallel)
What happens if only its length is doubled? (Hint: it's like having 2 original resistors in series)
What happens if you do both together?
 
  • #3
Vanadium 50 said:
Trying to do this in one bite.

What happens if only its diameter is doubled? (Hint: the area goes up by 4, so it's like having 4 original resistors in parallel)
What happens if only its length is doubled? (Hint: it's like having 2 original resistors in series)
What happens if you do both together?
(D) . Right ?
 
  • #4
Are you asking me or telling me?
(i.e. I'm not going to give you the answer.)
 
  • #5
Look at your calculations again. Is A2 = A1/4?
 
  • #6
mjc123 said:
Look at your calculations again. Is A2 = A1/4?
It should be A2 = A1*4
R = ##\frac{ρ*2l}{4A}##
R = ##\frac{ρl}{2A}##
R2 = R1 /2
 
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Related to Find the new resistance if its length and D are doubled

1. What is meant by "new resistance" in this context?

The new resistance refers to the electrical resistance of a material or circuit after its length and diameter (D) have been doubled. Resistance is a measure of how difficult it is for electricity to flow through a material, and it is affected by the length and thickness of the material.

2. How does doubling the length and diameter affect the resistance?

Doubling the length and diameter of a material or circuit will result in a four-fold increase in resistance. This is because the longer length and larger diameter provide more obstacles for the flow of electricity, making it more difficult for current to pass through.

3. Is there a formula for calculating the new resistance?

Yes, the formula for calculating the new resistance when length and diameter are doubled is R' = 4R, where R' is the new resistance and R is the original resistance. This formula is based on the relationship between resistance and length/diameter, which states that resistance is directly proportional to length and inversely proportional to diameter.

4. Can the new resistance be greater than the original resistance?

Yes, the new resistance can be greater than the original resistance if the original resistance was relatively low. For example, if the original resistance was 1 ohm, doubling the length and diameter would result in a new resistance of 4 ohms. However, if the original resistance was already high, doubling the length and diameter may not result in a significant increase in resistance.

5. Can the new resistance be calculated if only one of the parameters (length or diameter) is doubled?

Yes, the new resistance can still be calculated if only one of the parameters (length or diameter) is doubled. The formula for calculating the new resistance in this case would be R' = 2R, where R' is the new resistance and R is the original resistance. This is because doubling either length or diameter will result in a two-fold increase in resistance.

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