Question

In the figure below

Take R1= 50 Ω and R2 = 150 Ω. Suppose an ideal 28-V battery is connected across points A and B and, as a result, the power dissipation in R3 is 925 mW. Find all possible positive values for the resistance R3

لهم B.

Answer 1

R = 502 - I. R2 = 1501 & R₂ b=925mW - = 0.925W V-28V I ------------- Ha At node A, I, = I + Iz_- - using Kirchoff's voltage rule in loop 4 - -28 +I, R, + IL R2 = 0 - 3-28 +(4+13) x 50 + 12x150 Freud ca o 2 200 I, +5012 =28 = 100 I +2512=14:- - R and Raare in parallel so- Voltage across Ro'z voltage across Rz . R I = R₂ Iz = I= R3 Iz IN I= R₂ Iz 150 Using eq. 3 in 29. © - Hogy Rais + 25 13 = 14 > I 223 +75 = 14 2 Iz = 424 C2 R2+75) I3J

It is given that fewer dissipation by Rz is 0.9270 Il Rz = 0.925 - [ 42 px R g = 0.925 {using eq. } → -- 1764 R3 + It is given that fower dissipation by Rz is 0.925W -ot means IRg = 0.925 42 Pix Rg = 0.925 1 (2R3+75) 1764 R2 = 0.925 (2R2 +75)? 1764 R2 = 0.925/48 +5625 + 300 R3) » 34 8 – 14Q6.5 RA 15203-12S =0 This represents a leadratic egun in Rz. Its Coefficients are - a=3.7, b= 1486.5, C =5203.125 using quadratic eque formula - R₂ = -b ta b²-4ac za = 1486.5 I -14865)2-4837X5203.125 2x3.7 = (1486.57 1460.37) 7.4 on taking positive Sagna. Rg = 1486.3 +1460-37) = R3 = 398.231 on taking negative sign- R₂=3.532 7.4 = (1486.5-1460-37) = , 7.4