A certain number of identical resistors are connected in series and their total resistance is measured. When the resistors are disconnected and re-connected in parallel, their total resistance drops to one-percent of the value measured when they were connected in series. How many resistors are there?
If each resistor has value R, then
R/n = nR/100
100 = n^2
n=10
Where does 100 come from?
what, you've forgotten your algebra? Multiply by 100nto lear the fractions:
R/n = nR/100
100R = n^2R
100 = n^2
Steve, if you can, look at the answer given to me by Chegg. It is much more explanatory how 100 comes from than from what you wrote and just told me? 100 is not in the initial problem, so how am I to assume 100 is part of it? The only number given to me is 1%.
1% = 1/100
sheesh!
To solve this problem, let's assume that there are "n" identical resistors connected in series.
When resistors are connected in series, their resistances add up, so the total resistance in series is given by:
Total resistance in series = n * resistance of each resistor
Now, when the resistors are connected in parallel, their total resistance decreases. In parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistor's resistance. Therefore, the total resistance in parallel is given by:
1 / Total resistance in parallel = Sum of the reciprocals of the resistance of each resistor
In this case, the total resistance in parallel is said to be one-percent (1/100) of the value measured in series. So, we can write:
1 / Total resistance in parallel = (1 / 100) * (Total resistance in series)
Substituting the equations for the total resistance in series and parallel, we get:
1 / (n * resistance of each resistor) = (1 / 100) * (n * resistance of each resistor)
Simplifying the equation, we can cancel out the resistance of each resistor from both sides:
1 / n = 1 / 100
Now, we can solve for "n" by taking the reciprocal of both sides:
n = 100
Therefore, there are 100 resistors.