Question

Complete the lab 5
SERIES CIRCUITS VW REFERENCE READING Principles of Electric Circuits: Sectang 81 through 6-4. RELATED PROBLEMS FROM PRINCIPLES OF ELECTRIC CIRCUITS Chapter 5, Problems 1 through 26. OBJECTIVE To verify that the total resistance in a series-connected circuit is the sum of the individual resistances. EQUIPMENT dc power supply 0-10 V DMM or VOM Resistors (+5%): 110 2k12 322 3.9 kn 6.1 kn BACKGROUND When resistors are connected in series, the current that will flow is calculated from a knowledge of the total resistance. The total resistance in this case is simply the spm of the individual resistors, By knowing the total resistance, the voltage required for a desired current or the current resulting from an applied voltage can easily be calculated. One of the purposes of this experiment is to provide you with still more experience in using your DMM. You will use it on the ohms, volts, and milliampe functions in verifying the Series Resistor Rule.

Answer 1

Currently iam in home . I don't have any equipment like DMM. So, I am calculating manually.

=> Given circuit is, I aimat {R3 Ry Table 8-1:- 1) As per question, they give (+5%) tolerance 282 јко 13k2 3.9612 . 5.1k2 | Ry = 15k.2 El. 1850-12 = 14,2500 (06) :9502 :37052484522 (ov) : 3.705k4 4.845K -1900 (0) 119k.22 (oo) 2.85km (oo) = 0.95kr -15, 7501 1=31sor =40952 1:53552 (00) (0) ( = 15.75kn - lososa (oo -1.05 km 1 : 21002 = 2.1k = 4:095kn = 5.355kn 1=3.15ks Rmin Ranax IKS2 = 1000.2 = 5). X10001 - 50 ->1000- =950 =0.95kn Oko 20000 = 54. X 2000 = 10o =% 2000-100 =19002 = 1.9kn e5 X3000 - 150 -3000-150 = 2850-12 = 2.85K2 1000+ so - losos -1.askr = 2000+100 = 21002 22.1kr = 30006150 = 31502 = 3.15kr la => 3K1 - 3000.12 - 502 X3000 => 3.9 kur = 39001 - 50% X 3900 : 19 - 56 X 3900 - 195 → 3900-195 = 370512 1. = 3.7o5ka - 3900+195 = 40952 =4.095kr! 1 Inen 50a X5100 = 355 > 5.1ks =sloon = 56x5 = 5100-255 - 484532 -4.845K12 - 5100+255 -535512 = 51355kn => Ry = 15KR =15,0001 0.5kn =15,000_- 50% X 15,000 = 750 = 15,000-750 . =15,000+ AJO Ooh =1575an = 14,25002 = 14.25k2 -15.75 km

2) 73) FIESTAS I don't have any instrument to measure resistance. from Table; a Rt (min) - 14.25 km2 = 095+19K 72-85k+370.5k +4.945* 14.25 km2 = 14.25K2 and Rt (max) = 15.75kn = 1:05K+2JK +3•15k + 4.095545.355k 15.75kn = 15.75km ..Rg is equal to sum of individual resistances. 10 15K 4). Given, Vs = 10V = 0.67 mA > Nominal Cussent (ING → minimum Cursent (Imin) = V = 10 = 0.70lm A RE(min) 14.25% - maximum Cussent (Imax) = zimax at to us = 10 = 0.635mA 15.75k Table 8- 28- Inom (mA) | 0.67 Vs = 10V Imak (MA) 10.701 - Imagt (mA) 10.635 - Imean is measured using only by & connecting the circuit Dmm equipment as given. =>6) Given I=0.5mA > Nominal value of Vs = IRy = 0.5m *15% - 7.5U - minimum value of Us = I Ritmin) = 0.5m * 14.25K = 7.1250 maximum value of Vs = I max) = 0.5m x 15.75k = 7,875 v

7 Table 8-3: - | Unom (v) | 7.5* | I=0.5mA | Vmen cu) | 7.125 | Vmax(u) 7.875 CY . 7 I mean can be measured using Dmm. we mean * Questions: 1. If the measured value of Ry didn't equal the sum of the andividually measured resistances, you might conclude that sd am thinking the reason for above question 98 all of the above. ason

IQ YR 3) >> To calculate Imrin we need to 28 maximum resistance So, given tolerance +50% >> Because " For Iks, akr and 3k2 JIKA = Rmax = 1050.62 akn = Rmax = 21002 3K2 = Rmax = 3150 To ma :: Rt (max) = 6300 ... Imen = ex remax) - 6300 = 1.587 n 1.59M A

3) The cussent through each asistov in a sexies-connected (percuit such as that in figure 8-1 is the same I Sol True In a series - Connected resistor circuit, cussent is same for each aclistor as same. - 374) Given Given 10 %. tolerance. & Find maximum cussent, we consider $10% alks : 1000*100 = 100 = Presentation - (1000-100) - 900.2 > Okn - 2000 * 10= 200 =) Recept = (2000–200) = 1800-12 -73ks - 3000 floola = 500 -> Rovex = (3000 -300) = 27000 → 3.9K2 = 3900 * 10 4/0 - 390 = Roman = (3900 -370) = 35102 → 5.1kr = 5100 * 10°2 = 510 => Rimin = (5100-510) = 4590.2 n = 900t1800 + 2700 +3510 + 4590 = 13.5k2 i, Imax = Brin) = 0 = 0.74 mA W 740 MA :-50, option (C) as corect. 13.5k
5). If any resistance value reduced in series circuit, the total resistance will also reduced, then automatically current will increase.

Because , current and resistance are inversely proportional. So, minimum resistance will lead to maximum current.