Question

HELP IN PHYSICS PLEASE?

Answer 1

Consider two resistors in series R1 = 5 ohms and R2 = 7 ohms. They are powered with a battery providing a potential difference of 48 volts

Answer 2

How do you calculate the total current leaving source: I
- current across R1: I1
- current across R2:I2
the potential drop across R1:V1 and R2:V2

Answer 3

\[V=IR\]

Answer 4

The voltage is equal to the intensity of the current multiplied by resistance

Answer 5

But if im trying to find the total and theres a difference of 48 V
do i use tht?

Answer 6

through the R_1 the current is 48V/5ohms ~ 9.5Amps

Answer 7

alright so I = 9.5 all the way

Answer 8

So when it asks for the current across R1: I1
it would just be 9.5

Answer 9

Yeah actually 9.6 Amps through the first resistor

Answer 10

even when it asks the current across R2: I2

Answer 11

Throught the second resistor R_2 the current is
\[I=V/R\]
\[=I_2=48V/7\Omega\approx 6.9 Amps\]

Answer 12

oh alright . but how does the potential drop work?

Answer 13

nevermind the drop would be
V = 9.6 x 5
V = 6.9 x 7
right?

Answer 14

what happens when you add those together?
you should get a total of 48V

Answer 15

well i get 48 + 48.3 = 9.6 ...

Answer 16

i think i must have said something wrong because add those give a greater potential that then the battery supplied

Answer 17

ah ok i see what my mistake was , sorry about this,

The current in the circuit is \[I=V/(R_1+R_2)\]

Answer 18

So I = V / (R1 + R2)
would be the formula for finding the circuit of R1:I2

Answer 19

AND finding R2:I2

Answer 20

The current will be the same at all point in the circuit

\[I_T=\frac V R_T=48/(5+7)=4Amps\]

Answer 21

thats right two last questions sorry im terrible at physics.

Answer 22

how do you calculate the potential of the circuit at the point between the two resistors (R1 and before R2)

Answer 23

LAST- Calculating the potential after the current traverses R2 and why?

Answer 24

The current is constant through out the circuit,The voltage drop across R_1 will be 4x5=20V

Answer 25

Because the circuit the in series\[I=I_1=I_2\]

Answer 26

\[V_2=4 \times 7 =28V\]

Answer 27

So between the points the circuit stays the same.

Answer 28

yeah think of the flowing electrons, they arn't leaving the circuit at any point

Answer 29

how about calculating the traverses R2

Answer 30

I dont know what you mean by traverse

Answer 31

like calculating the potential after the current traverses R2

Answer 32

DO you mean what is the potential voltage across R2?

Answer 33

yeah lol

Answer 34

\[V_2=IR\]
\[=4A\times 7\Omega=28V\]

Answer 35

Oh alright thank you so much! (:

Answer 36

So \[V_1+V_2 =V\]
\[=20V+28V=48V\]

all the power int the battery has gone to the potential across the resistors

Answer 37

\[I=I_1=I_2 = 4A\] at all points in the (series)circuit the current is the same