Question

When connected in series to a 110-V line, two resistors use one-fourth the power that is used when they are connected in parallel. If one resistor is
2.0 kΩ, what is the resistance of the other?

Answer 1

so confused

Answer 2

draw a couple of circuit diagrams first

Answer 3

then write down expressions for the power consumed in each circuit, in terms of the supply voltage and the resistor values, just call them R1 and R2.
you know from the question that the ratio of these two expressions is 4 (or1/4 depending which you divide by which), and this will give you an expression relating the values of the two resistors, the supply voltage will cancel out

Answer 4

so for series its P=I^2 (R1 +R2)

and for parallel its P=V^2(R1 + R2) /(R1R2)

but from here is where I get confused as to what to do next

Answer 5

for the series expression, write it in the V^2/R form as well, don't introduce the current I

Answer 6

does that make sense ?

Answer 7

you know that the power in the parallel case is 4 times the power in the series case, so you can divide one expression by the other and get a relation between R1 and R2

Answer 8

so make it P=V^2(R1 + R2)?

Answer 9

I'm still struggling to see it

Answer 10

well in the series case it has to be V^2/(R1+R2)

Answer 11

and you know that P(parallel)/P(series)=4

Answer 12

so the V's cancel and I get

((R1 +R2)/R1R2) x R1R2=4

is that correct?

Answer 13

no, it should be x (R1+R2) , not x R1R2

Answer 14

so would they end up cancelling as well?

Answer 15

what do you mean by 'they' ?

Answer 16

((R1 +R2)/R1R2) x (R1 +R2) =4

so would the (R1 + R2) cancel leaving me with just R1R2?

Answer 17

no , the R1+R2 terms should both be on top

Answer 18

so with both on top would I then plug in my R1 value and attempt to solve R2?

Answer 19

You need to find the value of two resistors that when connected in series is 4 times their value when they are connected in parallel. Well any pair of equal resistors when connected in parallel is 1/2 their value, and when connected in series is twice their value or 2R/(1/2R) = 4 But out of all the infinite possibilities they have selected a resistor of 2K for one resistor, now you will have to select the other resistor. Make your selection equal to the problems selection.

Answer 20

You can then prove your selection by working out the power and showing the ratio of 4.

Answer 21

Show that :