OpenStudy (mtalhahassan2):

I need help

1 year ago
OpenStudy (mtalhahassan2):

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

1 year ago
OpenStudy (mtalhahassan2):

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?

1 year ago
OpenStudy (mtalhahassan2):

@Astrophysics

1 year ago
OpenStudy (mtalhahassan2):

@jackthegreatest

1 year ago
OpenStudy (mtalhahassan2):

@SapphireMoon @razor99 @likeabossssssss @RedR0se03399 @Bigbosssaint21

1 year ago
OpenStudy (mtalhahassan2):

@3mar

1 year ago
OpenStudy (sapphiremoon):

Sorru, I don't know resistance.

1 year ago
OpenStudy (sapphiremoon):

*Sorry

1 year ago
likeabossssssss (likeabossssssss):

sorry

1 year ago
OpenStudy (mtalhahassan2):

@Austin.L

1 year ago
OpenStudy (mtalhahassan2):

@Directrix

1 year ago
OpenStudy (mtalhahassan2):

can someone able be to help me out

1 year ago
OpenStudy (mtalhahassan2):

@Whitemonsterbunny17

1 year ago
OpenStudy (mtalhahassan2):

@DanJS

1 year ago
OpenStudy (mtalhahassan2):

@rishavraj @Nnesha @Hayhayz @Luigi0210

1 year ago
OpenStudy (3mar):

Sorry, I was not online!

1 year ago
OpenStudy (3mar):

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!

1 year ago