Question

I need help

Answer 1

If R1 = 5 kΩ, R2 = 9 kΩ, R3 = 7 kΩ, and DC = 15 V; what is the voltage drop seen across R2?

Answer 2

http://prnt.sc/cv79q1

Answer 3

do you agree?

R_total = 5000 + 1/(1/9000 + 1/7000)
R_total = 8937.5 ohms

use Ohms law
V=I R15=I∗8937.5I=15/8937.5I=0.001678321678 A

I got up to here but now how can I find the voltage drop across R2?

Answer 4

@Astrophysics

Answer 5

@jackthegreatest

Answer 6

@SapphireMoon @razor99 @likeabossssssss @RedR0se03399 @Bigbosssaint21

Answer 7

@3mar

Answer 8

Sorru, I don't know resistance.

Answer 9

*Sorry

Answer 10

sorry

Answer 11

@Austin.L

Answer 12

@Directrix

Answer 13

can someone able be to help me out

Answer 14

@Whitemonsterbunny17

Answer 15

@DanJS

Answer 16

@rishavraj @Nnesha @Hayhayz @Luigi0210

Answer 17

Sorry, I was not online!

Answer 18

WHat you get is correct and needed for the next step, which is: finding the voltage drop across R1.
\[V_{R_1}=I*R_1=(0.001678)*(5000)=8.39 V\]
And as you know that the voltage of the source is equal to the sum of the voltages across the series resistances, i.e
\[V_{source}=V_{R_1}+V_{R_2,R_3}\]

So it is obvious now to determine the voltage drop seen across R2:
\[V_{R_2}=V_{R_3}=V_{source}-V_{R_1}=15-8.39=6.61V\]

I hope that helps
and
Sorry again for being late!