electricity question

[Oh_Gee]

from GS-8

37. R1 and RT are both in series so while i understand that the resistance will be bigger in RT giving it a bigger voltage drop and R1 a smaller voltage drop, but shouldn't current be the same because it is series?

39. I tried to use P= (I^2)R
R is given by the table and I found I by doing I=2/480 but that didn't work. i don't understand their equations

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[3655]

37. Because both resistors are in series, they will have the same current so since Rt will be bigger and the total resistance will be bigger while the voltage is held constant then the current for the whole circuit will be smaller. The current in RT will also be the current in R1 since they are in series. So since the R1 is held constant while the current is smaller, the voltage in R1 will be smaller.

39. Remember since the resistors are both in series, you have to add both resistors to get the current so I=2/480+1000.

Hope that helps.

 

[Oh_Gee]

37. Because both resistors are in series, they will have the same current so since Rt will be bigger and the total resistance will be bigger while the voltage is held constant then the current for the whole circuit will be smaller. The current in RT will also be the current in R1 since they are in series. So since the R1 is held constant while the current is smaller, the voltage in R1 will be smaller.

39. Remember since the resistors are both in series, you have to add both resistors to get the current so I=2/480+1000.

Hope that helps.

37. wait why do you say "current is smaller," if the current is the same across both resistors

39. so you did I=V/Req which is for series. i get that but what equation are they using in the explanation?


thanks in advance

 

[3655]

37. Since the RT is bigger, it makes the total resistance bigger and since voltage is the same, the current across the whole circuit will be smaller. The current in RT will be small as well since RT and R1 are in series. So V=IR. Req is equal to 1000 +480= 1480. V is not changed so it is still 2 so now you got 2=I(1480), so I is 2/1480~(1.35 *10^-3) . The I that you get from this will be the new current across the circuit and since the resistors are in series, the current (1.35 *10^-3) will be the same for RT and R1. Since the new current is smaller than usual, R1 is still 1000 but multiplied by a smaller I which will result in a smaller voltage.

38. The equation for power they used is p= v^2/R. So they used voltage to find the Power instead of current, however you can still use current to find power as well like you did.