18) rezistenta interioara=r
P ext = RE patrat/ (R+r) patrat
[tex]p \: ext = \frac{5 {e}^{2} }{ {(5 + r)}^{2} } [/tex]
[tex]p \: ext = \frac{0.2 {e}^{2} }{ {(0.2 + r)}^{2} } [/tex]
Egalam cele doua puteri exterioare:
[tex] \frac{5 {e}^{2} }{ {(5 + r)}^{2} } = \frac{0.2 {e}^{2} }{ {(0.2 + r)}^{2} } [/tex]
[tex] {(5 + r)}^{2} \times 0.2 {e}^{2} = {(0.2 + r)}^{2} \times 5 {e}^{2} [/tex]
Impartim prin e patrat:
[tex] {(5 + r)}^{2} \times 0.2 = {(0.2 + r)}^{2} \times 5[/tex]
Impartim totul la 0.2:
[tex] {(5 + r)}^{2} = {(0.2 + r)}^{2} \times 25[/tex]
Desfacem parantezele:
[tex] {r}^{2} + 10r + 25 = ( {r}^{2} + 0.4r + 0.04) \times 25[/tex]
[tex] {r}^{2} + 10r + 25 = 25 {r}^{2} + 10r + 1[/tex]
-r patrat
10r+25=24r patrat+10r+1/-10r
25=24 r patrat+1/-1
24= 24 r patrat/ ÷24
r patrat = 1 => r= 1 ohm
