Supplementary Examples on Resistance and Application, Assignments of Electrical Engineering

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Supplementary Examples #2 Resistance and Application Supplementary Example 3.1 With reference to the figure below, calculate the following. 3.1.1 The total resistance (383.2Ω) 3.1.2 The current I3 (use current divider rule) (9V) 3.1.3 The voltage drop across R3 (use voltage divider rule) (7.916V) 3.1.4 If all the resistors in the circuit had a tolerance of ±5%, determine their colour code. 3.1.1 Ω= += +=− 99 220 1 180 1 11 1// 21 1 1// R RR R Ω= += +=− 41 82 1 82 1 11 2// 54 1 2// R RR R B A R1 =180Ω R7=1200Ω I3 R2 =220Ω R3 =330Ω R4 =82Ω R5 =82Ω R6 =15Ω E =12 V Ω= += += 1215 120015 761 RRRSer Ω= += += 371 41330 2//32 RRRSer The equivalent circuit is shown in the figure below Ω= += +=− 2.284 371 1 1215 1 11 3// 21 1 3// R RR R SerSer Ω= += += 2.383 2.28499 3//1// RRRT A B R//1 =99Ω RSer2 =371Ω E =12 V RSer1 =1215Ω A B R//1 =99Ω R//3 =284.2Ω E =12 V Ω= + × = + = += 76.300 57074.636 57074.636 111 21 21 2 212 SS SS P SSP RR RRR RRR mA R EI rRR RRR T T ST PS 42.13 76.521 7 76.521 176.520 76.520 22076.300 3 623 = = = Ω= += += Ω= += += 3.2.2 V VEV mV rIV rT Tr 99.6 1042.137 42.13 11042.13 3 3 = ×−= −= = ××= ×= − − 3.2.3 R1 - 100Ω - Brown, Black, Brown, Gold R2 - 330Ω - Orange, Orange, Brown, Gold R3 - 560Ω - Green, Blue, Brown, Gold R4 - 470Ω - Yellow, Violet Brown, Gold R5 - 100Ω - Brown, Black, Brown, Gold R6 - 220Ω - Red, Red, Brown, Gold Supplementary Example 3.3 The coil of a relay takes a current of 0.12A when it is at the room temperature of 15°C and connected across a 60V supply. If the minimum operating current of the relay is 0.1A, calculate the temperature above which the relay will fail to operate when connected to the same supply. Resistance temperature coefficient of the coil material is 0.0043 per °C at 0°C. Solution: Resistance of the relay coil at 15°C is… Ω= = 500 12.0 60 15R Let t2 be the temperature at which the minimum operating current of 100mA flows in the relay coil. Ω= = 600 1.0 60 FailR Therefore 1...)1( 150015 equtRR α+×= and 2...)1( 200 equtRRFail α+×= From …equ 1 and …equ 2: Ct where t tR tR R R Fail °= ×+ ×+ = + + = 51.64 ])0043.0[1( ])150043.0[1( 600 500 )1( )1( 2 2 200 150015 α α If the temperature rises of the coil rises above the value of t2, the increase in resistance then due to temperature co-efficiency of metals, the relay coil will draw a current less than 0.1A and, therefore, will fail to operate. Supplementary Example 3.4 A short overhead D.C. distribution line of 2km in length has a copper conductor with a diameter of 10mm which is suitable to carry 50A. The current carrying capacity of the distribution line must however be increased to 200A. Calculate the diameter of an aluminium conductor which can be connected in parallel with copper conductor. Take the resistivity of copper to be 0.017µΩ and that of aluminium to be 0.028µΩ.m ( ) 26 23 2 1054.78 1010 4 4 m daCu − − ×= ××=       = π π ( ) Ω= × ××= = − − 4329.0 1054.78 200010017.0 6 6 a RCu ρ HINT: The copper conductor is carrying 50A at the above calculated resistance and the aluminium conductor is carrying 150A with unknown resistance. An aluminium conductor of the same length will have 3 times less resistance because it carries three times less current. Ω= = 1443.0 3 4329.0 AR ( ) 26 6 101.388 1443.0 200010028.0 m R aA − − ×= ××= =   ρ mm m ad A 26.22 1026.22 109.3884 4 3 6 = ×= ×× = × = − − π π  3.6.1 Determine the colour code values of the resistors below 3.6.2 Determine the value of the total resistance using resistor colour codes [797Ω] 3.6.3 Calculate the value of I4 flowing through resister R4 in figure 6 above by applying the current divider rule. [7.466mA] 3.6.4 Determine the potential difference across R3 in figure below by applying the voltage divider rule. [7.466V] 3.6.1 %51200R %2220R %201000R %5180R %1012R 5 4 3 2 1 ±Ω= ±Ω= ±Ω= ±Ω= ±Ω= 3.6.2 ( ) ( )[ ] Ω= ++= Ω= ++ ×+ = 797 605 2//1 543 543 // RRRR RRR RRRR T 3.6.3 ( )[ ] mA RRR RII mA R EI T 466.7 06.15 543 5 4 = ++ ×= = = 3.6.4 ( ) ( )[ ] V II RIRIEVR 108.9 180121006.1512 3 21 22115 = +××−= = +−= − R1 I2 R5 I5 I1 R2 R4 R3 I3 I4 E = 12V V RR RVV RR 466.7 2201000 1000108.9 43 3 53 = + ×= + ×= Direct Current Distribution Often a source of electrical power is not situated close to where the power is needed and thus needs to be transported from one point to another. A large distance between source and destination is not necessarily such a bad thing since not many people would like to live next to a coal burning power station. However, when electricity has to be transported over large distances losses occur in the transmission cable. This has to be minimised as this power is lost or wasted. Figure 3.7.1 We will only look at D.C. distribution, but we will still see the effect of losses in the transmission cable. Figure 3.7 shows a very basic D.C. distribution system. On the left-hand side is the source of electrical power and on the right hand side is the load or device which will use the electrical power. Joining the source to the load are the positive and negative distributors where the losses occur. Power losses in the transmission cable can be calculated using any of the following formulae for power. RI VI R VPLoss 2 2 = = = In figure 3.7.2 we have an example of a D.C. distribution system. A 400V supply is feeding a load which is 200m away with a current of 50A. If the resistance of the cable used is 0.001Ω per metre, calculate the efficiency of the system. Negative Distributor LOAD = 20A + - Supply Positive Distributor D.C. DISTRIBUTION SYSTEM Figure 3.7.2 We will first calculate the resistance of the cable. Total cable length is… m400 2200 = ×= If 1 meter has a resistance of 0.001Ω then 400m will have a resistance of Ω= ×= 4.0 001.0400R If W VIPIN 20000 50400 = ×= = W RI PPP LossINOUT 19000 )4.050(20000 )(20000 2 2 = ×−= −= −= %95 100 20000 19000 %100 = ×= ×= IN OUT P Pη LOAD = 50A + - 400V 200m 3.7.1 V V A 4.230 240 %100 %96 160 = ×= V VV EFAB 8.4 2 4.230240 = − = = Ω= = = = 0133.0 360 8.4 I V RR AB EFAB Ω= ×= = 008.0 0133.0 250 150 DEBC RR 2 6 376 0133.0 2501002.0 mm R a a R AB AB AB AB = ××= = = −   ρ ρ 240V LOAD 2 = 200A A F LOAD 1 = 160A 250m 150m E B D C 360A 200A 3.7.2 𝑃𝑃𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 𝐼𝐼2(𝑅𝑅𝐴𝐴𝐴𝐴 + 𝑅𝑅𝐸𝐸𝐸𝐸) + 𝐼𝐼22(𝑅𝑅𝐴𝐴𝐵𝐵 + 𝑅𝑅𝐷𝐷𝐸𝐸) = [(3602 × 0.0266) + (2002 × 0.016)] = 3447.4 + 640 = 4087.36𝑊𝑊 𝐸𝐸𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = 𝑃𝑃𝑃𝑃 = 4.087𝑘𝑘𝑊𝑊 × 6ℎ = 24.52𝑘𝑘𝑊𝑊ℎ 𝐶𝐶𝐶𝐶𝐶𝐶𝑃𝑃 = 𝐸𝐸𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 × 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 24.52𝑘𝑘𝑊𝑊ℎ × 𝑅𝑅1.75/𝑘𝑘𝑊𝑊ℎ = 𝑅𝑅42.91