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I would like to find the power absorbed by the voltage source B. I know that \$P=VI=RI^2=(0.5\Omega)\left( \frac{4-E}{1\Omega + 0.5\Omega}\right)^2\$.

What is the "right way" to find it?

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Edit: Basically it gives me three options: \$E=1[V], E=0[V], E=-2[V]\$, and I need to find \$E\$ such that B absorbs the maximum power. Textbook's solution is \$E=1[V]\$.

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  • \$\begingroup\$ Other solution???? \$\endgroup\$
    – Andy aka
    Jan 22, 2020 at 13:28
  • \$\begingroup\$ What solution does the textbook give? \$\endgroup\$
    – Huisman
    Jan 22, 2020 at 13:29
  • \$\begingroup\$ Your power equations are incorrect. hyperphysics.phy-astr.gsu.edu/hbase/electric/elepow.html \$\endgroup\$ Jan 22, 2020 at 13:36
  • \$\begingroup\$ @PeterSmith They are expressed in peak voltage and peak current. \$\endgroup\$
    – Kevin
    Jan 22, 2020 at 13:37
  • \$\begingroup\$ You should do the superposition theorem, once you have the mid-point voltage you will be able to measure the voltage difference across the resistor 0.5 ohm, then naturally you can measure the power absorbed. \$\endgroup\$
    – Delphesk
    Jan 22, 2020 at 13:37

2 Answers 2

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The power equation you start with is the the power absorbed by the resistor 0.5 ohm, but you should take into account the power absorbed by the voltage source as well:

\$P=R\cdot I^2+E\cdot I = 0.5\Omega\left( \frac{4\text{ V}-E}{1\Omega + 0.5\Omega}\right)^2 + E\left( \frac{4\text{ V}-E}{1\Omega + 0.5\Omega}\right) \$

\$ P = \frac{2}{9\text{ }\Omega} \left( 16\text{ V}^2-8E+E^2 \right) +\frac{2}{9\text{ V}\Omega}3 \left( 4E-E^2 \right) = \frac{2}{9\text{ }\Omega} \left( 16\text{ V}^2+4E-2E^2 \right) \$

\$ P = \frac{4}{9\text{ }\Omega} \left( 8\text{ V}^2+2E-E^2 \right) \$

Note you need to find \$E\$ such that B absorbs the maximum power, so P should be positive.

Decent way is to find optimum for E, so \$\frac{dP}{dE}=0\$

$$\frac{dP}{dE} = \left( \frac{4}{9\text{ }\Omega} \right) \frac{d}{dE} \Bigg(8\text{ V}^2+2E-E^2 \Bigg) = \left( \frac{4}{9\text{ }\Omega} \right) \left( 2-2E \right) $$

\$\frac{dP}{dE}=0\$ for \$E=1\$

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  • \$\begingroup\$ The power is \$I^2R+EI\$. Both the ideal voltage source and the 0.5 resistor are recipients of power. Your equation does not have units of power. \$\endgroup\$
    – Chu
    Jan 22, 2020 at 14:49
  • \$\begingroup\$ After corrections, the equations are still wrong. \$\endgroup\$
    – Chu
    Jan 22, 2020 at 15:01
  • \$\begingroup\$ @Chu ah, I see: I copy/pasted to \$E \cdot I^2\$ instead of \$E \cdot I\$ and continued with that ... will update \$\endgroup\$
    – Huisman
    Jan 22, 2020 at 15:10
  • \$\begingroup\$ For quadratic equation, there is no need to do derivative to find the min/max. \$\endgroup\$
    – X J
    Jan 22, 2020 at 17:00
  • \$\begingroup\$ @XJ You're right. I mistakingly copied \$E \cdot I^2|$ and ended up with a 3rd order... for a second order, this is overdone, but I kept the approach since it is also valid for higher orders. \$\endgroup\$
    – Huisman
    Jan 22, 2020 at 21:35
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IB = (4V-E)/1.5Ω=(8V-2E)/3Ω

VB=E+IB*0.5Ω=E+(4V-E)/3=(2E+4V)/3

PB=IB*VB=(8V-2E)*(2E+4V)/9Ω <=4W

and it gets to the maximun when (8V-2E)=(2E+4V) (according to AM–GM inequality) => E=1V

Or, look (8V-2E)*(2E+4V) as a quadratic function of E and let it equal to zero to get the two roots, 4V and -2V. The mid point of these two roots, 1V, corresponds to the max point.

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  • \$\begingroup\$ I must admit your approach is faster,+1 for that \$\endgroup\$
    – Huisman
    Jan 22, 2020 at 21:36
  • \$\begingroup\$ @Huisman, Thanks! \$\endgroup\$
    – X J
    Jan 22, 2020 at 21:52

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