A piece of copper wire of resistance R is cut into 10 equal lengths. These parts are connected in parallel. How does the joint resistance of the parallel combination compare with the original resistance of the single wire?
No drop
If all these parts are connected in parallel, that would mean the new total resistance R" can be found as follows:
1/R" = n^2/R
R" = R/n^2
R" = R/10^2
R"= 0.01
To solve this problem, we need to consider the relationship between resistance and length for a wire.
The resistance of a wire is directly proportional to its length. So, if the original wire had a resistance of R, each of the smaller wires will also have a resistance of R/10 since they are of equal length.
Now, when these smaller wires are connected in parallel, the total resistance of the combination is given by the formula:
1/R_total = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn
Since all the smaller wires have the same resistance (R/10), we can substitute this value into the formula:
1/R_total = 1/(R/10) + 1/(R/10) + 1/(R/10) + ... + 1/(R/10)
Simplifying the equation:
1/R_total = 10/(R/10)
1/R_total = (10/1) * (10/R)
1/R_total = 10^2/R
R_total = R/10^2
Therefore, the joint resistance of the parallel combination is 1/100 times the original resistance of the single wire.
To find the joint resistance of the parallel combination, we need to consider the formula for the equivalent resistance of resistors connected in parallel.
When resistors are connected in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances. Mathematically, this can be expressed as:
1/Req = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn
In this case, since the 10 equal lengths of copper wire are connected in parallel, we can apply this formula to find the joint resistance of the combination.
Let's say the resistance of each individual wire is R. Since the wire is cut into 10 equal lengths, the total number of wires is 10.
Now, let's substitute the values into the formula:
1/Req = 1/R + 1/R + 1/R + ... + 1/R (10 times)
Using the formula for the sum of reciprocals of equal terms (1/R + 1/R + ... + 1/R n times = n/R):
1/Req = 10/R
To get the value of Req, we can take the reciprocal on both sides:
Req = R/10
Hence, the joint resistance of the parallel combination is one-tenth (1/10) of the resistance of the original single wire.
Therefore, we can conclude that the joint resistance of the parallel combination is smaller than the original resistance of the single wire.