anonymous
  • anonymous
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
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Michele_Laino
  • Michele_Laino
total resistance, of our circuit, is given by the subsequent formula: \[\large {R_{TOTAL}} = 8.5 + 3.2 + \frac{1}{{\frac{1}{{13}} + \frac{1}{{18}}}} = ...\]
anonymous
  • anonymous
ok! we get 19.248?

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Michele_Laino
  • Michele_Laino
ok! So the current is: \[\large I = \frac{V}{{{R_{TOTAL}}}} = \frac{{15}}{{19.25}} = ...\]
anonymous
  • anonymous
ok! we get 0.779?
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
now 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|
anonymous
  • anonymous
5.889!
Michele_Laino
  • Michele_Laino
since, we have: \[{R_{AB}} = \frac{1}{{\frac{1}{{13}} + \frac{1}{{18}}}} = 7.55Ohm\]
Michele_Laino
  • Michele_Laino
\[\large {R_{AB}} = \frac{1}{{\frac{1}{{13}} + \frac{1}{{18}}}} = 7.55Ohm\]
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
what happens next then?
Michele_Laino
  • Michele_Laino
the 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
  • anonymous
1.9266? choice A is our solution?
Michele_Laino
  • Michele_Laino
yes! That's right!
anonymous
  • anonymous
yay! thank you! onto the next:)
Michele_Laino
  • Michele_Laino
:) ok!

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