A large spool in an electrician's workshop has 52 m of insulation-coated wire coiled around it. When the electrician connects a battery to the ends of the spooled wire, the resulting current is 2.3 A. Some weeks later, after cutting off various lengths of wire for use in repairs, the electrician finds that the spooled wire carries a 2.9-A current when the same battery is connected to it. What is the length of wire remaining on the spool?
ℰ=IR=IρL/A =>
I₁ρL₁A = I₂ρL₂A
L₂=I₁L₁/I₂=2.3•52/2.9=41.2 m
Well, it sounds like this electrician has been a real cut-up! The poor spool of wire has been unravelled, snipped, and spliced all over the place. But fear not, for I shall assist you with this electrifying problem!
Let's call the length of wire the electrician has cut off as "x". Initially, we are told that the total length of the wire on the spool is 52 meters. So, after cutting off "x" meters of wire, the remaining length on the spool is 52 - x meters.
Now, let's tackle the different currents. Initially, when the electrician connects the battery, the current is 2.3 A. After some time, with the same battery, the current becomes 2.9 A. We can use Ohm's Law to help us out here!
According to Ohm's Law, current (I) is equal to voltage (V) divided by resistance (R). Since the battery voltage remains constant in this scenario, we can say that the ratios of resistance and current are equal.
So, when the current is 2.3 A, the resistance (R₁) is equal to the voltage (V) divided by 2.3 A. Similarly, when the current is 2.9 A, the resistance (R₂) is equal to the voltage (V) divided by 2.9 A.
Since resistance (R) is directly proportional to length, we can write the following equation:
R₁ / R₂ = (52 - x) / x.
Now, we know that the ratios of resistance are equal, so we can solve for x. Let's get crackin'!
R₁ / R₂ = (52 - x) / x.
(V / 2.3) / (V / 2.9) = (52 - x) / x.
Cross-multiplying yields:
(V / 2.3) * (x / V) = (52 - x) / x.
Simplifying further:
2.9x = 2.3(52 - x).
2.9x = 119.6 - 2.3x.
5.2x = 119.6.
Dividing both sides by 5.2, we find:
x = 119.6 / 5.2.
x ≈ 23 meters.
So, the electrician has cut off approximately 23 meters of wire.
Now, to find the remaining length on the spool, we subtract the cut-off length from the original length:
52 - 23 = 29 meters.
Ta-da! The length of wire remaining on the spool is 29 meters. I hope I've managed to weave through this problem with the utmost precision and humor!
To solve this problem, we can use the fact that the current flowing through a wire is directly proportional to its length. Therefore, we can set up a proportion to find the length of wire remaining on the spool.
Let L be the length of wire remaining on the spool.
We know that when the current is 2.3 A, the wire length is 52 m. Let's call this length L1.
We also know that when the current is 2.9 A, the wire length is L. Let's call this length L2.
Using the proportion:
(L1/L) = (I1/I2)
Substituting the given values:
(52/L) = (2.3/2.9)
Cross-multiplying, we get:
2.9 * 52 = 2.3 * L
Simplifying:
150.8 = 2.3 * L
Dividing both sides by 2.3:
L = 150.8 / 2.3
Calculating this value, we find:
L ≈ 65.9 m
Therefore, the length of wire remaining on the spool is approximately 65.9 meters.
To find the length of wire remaining on the spool, we need to calculate the length of wire that has been used for repairs.
Let's assume that x meters of wire have been cut off from the spool for repairs. This means that the remaining length of wire on the spool is (52 - x) meters.
We can set up a proportion using the currents and the lengths of wire. The current is directly proportional to the length of wire, so we have:
2.3 A / (52 m) = 2.9 A / x
To solve for x, we can cross-multiply and then divide:
2.3 A * x = 2.9 A * (52 m)
2.3x = 150.8
x = 150.8 / 2.3
x ≈ 65.57
Therefore, approximately 65.57 meters of wire have been cut off for repairs. The length of wire remaining on the spool is (52 - 65.57) or approximately -13.57 meters.
Since it is not possible to have a negative length of wire remaining, we conclude that there is no wire remaining on the spool.