## anonymous one year ago For the circuit drawn inside, what will be the voltage drop across R_2? A. 9.47 V B. 14.2 V C. 21.3 V D. 45.0 V

1. anonymous

|dw:1433873690686:dw|

2. johnweldon1993

When you combine all of the resistors...what do you get for the equivalent resistance?

3. anonymous

95?

4. johnweldon1993

Correct... so if we were to redraw our circuit now... |dw:1433874008656:dw| What would be the total current going through the resistor?

5. anonymous

45?

6. johnweldon1993

Current, not Voltage :) $\large I = \frac{V}{R}$

7. anonymous

oh oops! so 45/95? = 0.47? :/

8. johnweldon1993

Right...that is our current that goes through our entire circuit here Now that we know that, we break apart our circuit again to the original with the current drawn in |dw:1433874182301:dw|

9. anonymous

ooh so what happens next?

10. johnweldon1993

So now focus on R_2 |dw:1433874327033:dw| The voltage across R_2 would be $\large V_{R_2} = I_{R_2}\times R_2$ $\large V_{R_2} = 0.47A \times 30\Omega = ?$

11. anonymous

14.1? so choice B (14.2 V) is our solution?

12. Michele_Laino

that's right!

13. johnweldon1993

Indeed! Sorry walked away for a minute! lol

14. anonymous

haha yay! thanks!! :D

15. Michele_Laino

:)