Question

Two resistors connected in series have an equivalent resistance of 724.5 Ω, when they are connected in parallel, their equivalent resistance is 135.2 Ω. Find the resistance of each resistor.

Answer 1

544.65 ohm and 179.84 ohm

Step-by-step explanation:

Hello

Let

Resistor 1 (R1) and Resistor 2 (R2)

in series
`R_(eq) =R1+R2`

in parallel

`(1)/(R_(eqp) ) =(1)/(R1) +(1)/(R2)\\\\R1=R_(eqs) -R2\\\\R1 =724.5-R2 (equation 1)\\ R_(eqp)}=(R1*R2)/(R1+R2)\\\\\ 135.2 = (R1*R2)/(R1+R2)(equation 2)\\\\\\\\replacing 1 in \ 2\\\\probema data`

`135.2=((724.5-R2)(R2))/(724.5-R2+R2) }\\135.2=((-R_(2) ^(2)+724.5R2) )/(724.5)\\ 135.2*724.5=-R_(2) ^(2)+724.5R_(2)\\\\R_(2) ^(2) -724.5R_(2)  +97952.4 =0\\R_(2) =\frac{724.5 \±\sqrt{(-724.5^(2))-4(1)(97952.4) } }{2(1)}\\R_(2) =(724.5\±√((133092.25) ) )/(2(1))\\\\R_(2) =(724.5+364.81810)/(2) \\R_(2) =544.65 ohm\\\\R_(2) =(724.5-364.81810)/(2) =179.84 ohm`\\\\

let R2=544.65 and replace in equation 1

R1=724.5-544.65

R1=179.85

so, the resistors are 544.65 ohm and 179.84 ohm

Have a great day