Question

Determine the resistance of a resistor which must be placed in series with a 75 ohm resistor across 120V source in order to limit the power dissipation in the 75 ohm resistor to 90 watts...

Answer 1

Let that added resistance be x
Current in the circuit if x is added=>

I = V/Rtotal
I = 120 / 75+x
Power accros 75 ohm resistor
P = I2R
90 = (120/75+x)2 x 75

x2 + 5625 + 150x= 12000

x2 + 150x - 6375 = 0
x=34.6

Answer 2

ⓗⓔⓨ ⓜⓐⓣⓔ ! ___________________________________________________________________________

resistance are connected in series so,

equivalent resistance = (R + 75)ohm

now, find current

I= V/R

I = V/R+75

as , we know in series current is same

so,

power = V.I

current will be same but voltage differ. in series. .

90= V' * V/ R+75

V'= voltage along 75ohm resistance

apply v'= I*R

now,

90 = 120*120*75/(R+75)^2

(75+R)^2=40*4*75

5627+R^2+105R=40*4*75

R^2 + 150R = 6373

by solving this quadritic eqn u will get the value of R = 34.6ohm

ⓗⓞⓟⓔ  ⓘⓣ  ⓦⓘⓛⓛ  ⓗⓔⓛⓟ  ⓨⓞⓤ !

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