Question

Table 3 (5 points) AV1_5.0 V R(22) im(A) (A) % Error R1 100 R2 200 0.005 R3 ! 3300 0.006 R400 .OZU R5 w 200 0.005 3. Using Kirchhoff's Rules, construct enough mathematically independent equations to solve for the current of each resistor. Then calculate the % error between your measured and theoretical values for the current of each resistor. You MUST use Kirchhoff's Rules and show work to receive any credit. (15 points) 3R5 NAV Rz

Answer 1

Given, Ri=1000 R = 2008 R₂ = 3300 Ros 1000 R = 2004 . & R 24. - E=5v i Els 3 &h=5v Vw dot the current the flowing the through the loob , & and ③ are is, in and is respectively Apply xvL at the top o, we get ki + R311 - 12) = & > (R1 + R3) ij - Rzle = { 7 (100+320) i, - 320 12 = 5 7 30 i, - 330i2 a5 & 861 – 66 12 = 1 – li Apply qul at the loof 6 , we gerer R3l 12 - )) + Rq (12-13) -0. » (Rz tRq) iz – R311 - Rq iz to

> -3301, + 430 12 - 1001z = > -334, + 1312 - 13 - - we Apply Kuh at the wrofe , we get Relis-in) + (R5 +R,)iz - - Rq 12 + (R2 + R pt (6) iz to » rooiz + ( 200+100+ 200)13 -- -10012 + 500 iz > -i +siz -o (1) * From equation (w), is = 2/5 - (in) Putting the value of is in equation (ii) , we get - 331, + 13 12 - 10x z ao -331, +4212-22 20 → -33, + 4liy (u mutiplying equation 33 and equation (v) by 86, we get- 28 38 i, - 2178 i = 33 - 28 381, +3526i2 = 0 -13 18 12 = 33 iz = 0.02.4 Addin

Putting the name in in equation (iv), we get iz= 1 = 0. on = 0.005 A z = 0.005 a Puttings the value of in in equation (w), we get 331 = 412 i & 4180.029 33 = 0.02 A :. , = 0.03 A 12 = 0.028 A iz = 0.005A current through resistance & Ip = 0.03 A = 0.005 A I = 0.03 -0.029 = -0.006 FR = - iz = 0.029 -0.005 = 0.019 A = 0.005A Ils
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