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Question-81005




Question Number 81005 by mathocean1 last updated on 08/Feb/20
Answered by Rio Michael last updated on 09/Feb/20
V_r  =230 − 180 = 50 V  R_t  = ((5r)/(5 +r))  from the max power theorem,r = R  ⇒ r = 5 Ω  I_1  = (V_r /(R_t  )) = 20A  I_2 = (V/(R_(t2)  )) = 48A (approx)  I = I_1  + I_2  = 68A
$${V}_{{r}} \:=\mathrm{230}\:−\:\mathrm{180}\:=\:\mathrm{50}\:{V} \\ $$$${R}_{{t}} \:=\:\frac{\mathrm{5}{r}}{\mathrm{5}\:+{r}} \\ $$$${from}\:{the}\:{max}\:{power}\:{theorem},{r}\:=\:{R} \\ $$$$\Rightarrow\:{r}\:=\:\mathrm{5}\:\Omega \\ $$$${I}_{\mathrm{1}} \:=\:\frac{{V}_{{r}} }{{R}_{{t}} \:}\:=\:\mathrm{20}{A} \\ $$$${I}_{\mathrm{2}} =\:\frac{{V}}{{R}_{{t}\mathrm{2}} \:}\:=\:\mathrm{48}{A}\:\left({approx}\right) \\ $$$${I}\:=\:{I}_{\mathrm{1}} \:+\:{I}_{\mathrm{2}} \:=\:\mathrm{68}{A} \\ $$
Commented by mathocean1 last updated on 12/Feb/20
please sirs what means approx ?
$${please}\:{sirs}\:{what}\:{means}\:{approx}\:? \\ $$
Commented by Rio Michael last updated on 12/Feb/20
approximately
$${approximately} \\ $$

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