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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
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433873690686:dw

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2When you combine all of the resistors...what do you get for the equivalent resistance?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Correct... so if we were to redraw our circuit now... dw:1433874008656:dw What would be the total current going through the resistor?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Current, not Voltage :) \[\large I = \frac{V}{R}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh oops! so 45/95? = 0.47? :/

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Right...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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh so what happens next?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2So now focus on R_2 dw:1433874327033:dw The voltage across R_2 would be \[\large V_{R_2} = I_{R_2}\times R_2\] \[\large V_{R_2} = 0.47A \times 30\Omega = ? \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.014.1? so choice B (14.2 V) is our solution?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Indeed! Sorry walked away for a minute! lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha yay! thanks!! :D
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