Question

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

Answer 1

total resistance, of our circuit, is given by the subsequent formula:
\[\large {R_{TOTAL}} = 8.5 + 3.2 + \frac{1}{{\frac{1}{{13}} + \frac{1}{{18}}}} = ...\]

Answer 2

ok! we get 19.248?

Answer 3

ok! So the current is:
\[\large I = \frac{V}{{{R_{TOTAL}}}} = \frac{{15}}{{19.25}} = ...\]

Answer 4

ok! we get 0.779?

Answer 5

ok!

Answer 6

now the voltage drop across the parallel of the two resistances is:
\[\large {V_{AB}} = {R_{AB}}I = 7.55 \times 0.78 = ...\]

Answer 7

5.889!

Answer 8

since, we have:
\[{R_{AB}} = \frac{1}{{\frac{1}{{13}} + \frac{1}{{18}}}} = 7.55Ohm\]

Answer 9

\[\large {R_{AB}} = \frac{1}{{\frac{1}{{13}} + \frac{1}{{18}}}} = 7.55Ohm\]

Answer 10

ok!

Answer 11

what happens next then?

Answer 12

the requested power W_3 is given by the subsequent computation:
\[\large {W_3} = \frac{{{{\left( {{V_{AB}}} \right)}^2}}}{{{R_3}}} = \frac{{{{5.889}^2}}}{{18}} = ...watt\]

Answer 13

1.9266? choice A is our solution?

Answer 14

yes! That's right!

Answer 15

yay! thank you! onto the next:)

Answer 16

:) ok!