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anonymous
 one year ago
For the circuit shown, what is the power dissipated by R3?
A. 2.0 W
B. 2.5 W
C. 2.8 W
D. 5.2 W
anonymous
 one year ago
For the circuit shown, what is the power dissipated by R3? A. 2.0 W B. 2.5 W C. 2.8 W D. 5.2 W

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2total resistance, of our circuit, is given by the subsequent formula: \[\large {R_{TOTAL}} = 8.5 + 3.2 + \frac{1}{{\frac{1}{{13}} + \frac{1}{{18}}}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! So the current is: \[\large I = \frac{V}{{{R_{TOTAL}}}} = \frac{{15}}{{19.25}} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now the voltage drop across the parallel of the two resistances is: \[\large {V_{AB}} = {R_{AB}}I = 7.55 \times 0.78 = ...\] dw:1434310801168:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2since, we have: \[{R_{AB}} = \frac{1}{{\frac{1}{{13}} + \frac{1}{{18}}}} = 7.55Ohm\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2\[\large {R_{AB}} = \frac{1}{{\frac{1}{{13}} + \frac{1}{{18}}}} = 7.55Ohm\]

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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the requested power W_3 is given by the subsequent computation: \[\large {W_3} = \frac{{{{\left( {{V_{AB}}} \right)}^2}}}{{{R_3}}} = \frac{{{{5.889}^2}}}{{18}} = ...watt\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01.9266? choice A is our solution?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! That's right!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay! thank you! onto the next:)
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