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Show that "power supplied" = "power dissipated"
For the circuit of Fig. 3-42, show that the power supplied by the sources is equal to the power dissipated in the resistors.
Ans. [tex]P_T=940.3\text{ }W[/tex]
[PLAIN]http://img269.imageshack.us/img269/2010/dsc01005kn.jpg
[tex]P=VI[/tex]
The [tex]3\text{ }\Omega[/tex] resistor gets current [tex]I_{3\Omega}=\frac{500}{500+3}\times 10\text{ }A=9.94\text{ }A[/tex]
Correct so far? From this I get [tex]V_R=I_{3\Omega}\times R=29.82V[/tex] and also the power dissipated [tex]P_{3\Omega}=\frac{V_R^2}{R}=296.4\test{ }W[/tex]
So far so good?
Now I can also get the value for the dependent source:
[tex]3V_R=3\times 29.82\text{ }V=89.46V[/tex] ... to which [tex]1.75\text{ }V[/tex] is added for total of [tex]V_T=91.21\text{ }V[/tex].
The power dissipated on the large resistor is [tex]P_{500\Omega}=\frac{V_T^2}{R}=16.64W[/tex].
Since the "power dissipated" by the two resistors don't add up to [tex]P_T[/tex] (and sources have their polarities aligned), it is obvious I've done some mistake up to this point. What am I doing wrong?
Homework Statement
For the circuit of Fig. 3-42, show that the power supplied by the sources is equal to the power dissipated in the resistors.
Ans. [tex]P_T=940.3\text{ }W[/tex]
[PLAIN]http://img269.imageshack.us/img269/2010/dsc01005kn.jpg
Homework Equations
[tex]P=VI[/tex]
The Attempt at a Solution
The [tex]3\text{ }\Omega[/tex] resistor gets current [tex]I_{3\Omega}=\frac{500}{500+3}\times 10\text{ }A=9.94\text{ }A[/tex]
Correct so far? From this I get [tex]V_R=I_{3\Omega}\times R=29.82V[/tex] and also the power dissipated [tex]P_{3\Omega}=\frac{V_R^2}{R}=296.4\test{ }W[/tex]
So far so good?
Now I can also get the value for the dependent source:
[tex]3V_R=3\times 29.82\text{ }V=89.46V[/tex] ... to which [tex]1.75\text{ }V[/tex] is added for total of [tex]V_T=91.21\text{ }V[/tex].
The power dissipated on the large resistor is [tex]P_{500\Omega}=\frac{V_T^2}{R}=16.64W[/tex].
Since the "power dissipated" by the two resistors don't add up to [tex]P_T[/tex] (and sources have their polarities aligned), it is obvious I've done some mistake up to this point. What am I doing wrong?
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