# Problem №3 (15 points)

There is a current source with an internal resistance of $r=20$ Ohms. What external resistance should be connected to the source so that the power dissipated in the external resistance differs from the maximum possible by $25 \%$?

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# Solution:

Power of the current

$P=I^{2} R=\left(\frac{\varepsilon}{R+r}\right)^{2} R$

(2 points)

It will be maximum when $R=r$, i.e.:

$P_{\text {MAX }}=\left(\frac{\varepsilon}{R+r}\right)^{2} R=\left(\frac{\varepsilon}{20+20}\right)^{2} 20=\frac{\varepsilon^{2}}{80}$

(5 points)

The power in question differs by $25 \%$, i.e., $P=0.75 \cdot \frac{\varepsilon^{2}}{80}$.

We get:
$0.75 \cdot \frac{\varepsilon^{2}}{80}=\left(\frac{\varepsilon}{R+20}\right)^{2} R$

We obtain a quadratic equation, the roots of which are:

$R_{1}=6.7 \, \Omega$

$R_{2}=60 \, \Omega$