EECE 210 Lecture, Spring 2012

1 Chapter 3 Simple Resistive Circuits 2 Resistors in Series In a series connection of resistors, the same current flows through all the resistors, and voltages across the individual resistors may be added algebraically to obtain the total voltage across the series combination R1 R2 Rn + v2 – + v1 – vn + – i + – v + – v i 3 Using KVL v = v1 + v2 + … + vn. Using Ohm’s law v = Ri, we obtain v = R1i + R2i +…+ Rni = [R1 + R2 + … + Rn]i= Reqi R1 R2 Rn + v2 – + v1 – vn + – i + – v + – v i 4 In general   In general, if k resistors are connected in series, then Req = R1 + R2 +… + Rk 5 Resistors in Parallel  In a parallel connection of resistors, the same voltage exists across all the resistors, and the currents through the individual resistors may be added algebraically to obtain the total current 6 KCL  Using KCL, is= i1 + i2 + i3+ i4  But, , then 4 4 3 3 2 2 1 1 , and i i , i , R v R v R v R v i s s s s                      eq s s s R v R R R R v i 1 1 1 1 1 4 3 2 1 7 In General,  In general, for k resistors connected in parallel, we have  If k = 2: 2 1 2 1 2 1 ) ( R R R R R || R   8 Series-Parallel Connection, Example 1 Determine Req 9 Series-Parallel Connection, Example 2 10 Voltage Divider Rule Determine Vk ? 1. KCL is satisfied by having a common current i through all the resistors 2. From Ohm’s law: vk = Rki, and vSCR = (R1+ R2 +…+Rk)i vSRC R1 i R2 – + Rk Rn + + v1 v2 – – + vk – + vn – 11 Voltage Divider Rule Hence vSRC R1 i R2 – + Rk Rn + + v1 v2 – – + vk – + vn – n k k SRC k R R R R R v v       ... ... 2 1 12 Example: Voltage Divider Rule If a load RL is connected across R2, then vSRC R1 + + v1 R2 RL iL vL – + – + – – ) || ( ) || ( 2 1 2 L L SRC L R R R R R v v   13 Current Divider Rule – A resistive current divider, consists of two or more resistors in parallel. In the case of only two resistors, the current i divides into two components i1 and i2. From KCL: i = i1 + i2. – By assigning a common voltage v = vSRC across the three parallel elements, KVL is automatically satisfied around the two meshes, one consisting of vSRC and R1 and the other consisting of R1 and R2. In going around the mesh formed by R1 and R2, for example, there is a voltage rise vSRC across one of the resistors and an equal voltage drop vSRC in the other resistor. R1 R2 vSRC – + – + i1 i2 i 14 Current Divider Rule  From Ohm’s law: (1) where is the parallel resistance of R1 and R2. Dividing (1) by R1R2, we obtain (2) R1 R2 vSRC – + – + i1 i2 i i R R R i R i vSRC ) || ( 2 1 2 2 1 1    2 1 2 1 2 1 ) ( R R R R R || R   2 1 1 2 2 1 R R i R i R i    15 Current Divider Rule (2) From (2), we obtain 2 1 1 2 2 1 R R i R i R i    R1 R2 vSRC – + – + i1 i2 i 2 1 1 2 2 1 2 1 1 2 1 2 2 1 2 1 R R R i R i R R i i or R i R R i i R i       16 Example 17 Example: Find the voltage across the 4Ω resistor. 18 Example: Find the current through the 12Ω resistor. 19 Measurement of DC Currents and Voltages  Ammeters. An ammeter is an instrument that measures current through a circuit element when inserted in series with that element  Voltmeters. A voltmeter is an instrument that measures the voltage across a circuit element. Example 20 100 identical lamps rated at 12 V, 6 W each are to be connected across a 240 V supply such that the rated voltage of 12 V is applied to each lamp. Design an appropriate arrangement of the lamps and calculate the equivalent resistance, the total current drawn from the supply and the total power dissipated/generated in the circuit. 21 Solution: Since 240/12= 20, then 20 lamps may be connected in series across the supply. The voltage across each lamp will be the rated voltage of 12 V, assuming the lamp resistances are all equal. To accommodate 100 lamps, 5 such series combinations will have to be paralleled across the supply. The resistance of each lamp is R=v2/P=122/6=24 . The equivalent series resistance of 20 lamps is 2420 = 480. The equivalent parallel resistance of five of these series combinations is 480/5=96. The lamp current at 12 V is 6W/12V=0.5 A, which is also the current in each series combination. The total current drawn from the supply is 50.5 = 2.5 A. we may calculate P, the total power supplied: from the total current, P = 2402.5 = 600 W. From the total resistance, Psupplied =Pgenerated =2.52 96=600W . From the total number of lamps and their individual ratings: P = 1006 W = 600 W. 22 Measuring Resistance The Wheatstone Bridge  Many different circuit configuration are used to measure resistance  Here, we focus on just one; The Wheatstone Bridge which is used to precisely measure resistances of medium values, that is, in the range of 1 Ohm to 1 Mega Ohm.  In Commercial models of the Wheatstone Bridge, accuracies on the order of ± 0.1 % are possible. 23 The Wheatstone Bridge  The Wheatstone bridge is a useful circuit for determining the value of an unknown resistance Rx in terms of known resistances. R3 is varied until the ammeter current becomes zero, which is the condition for bridge balance. R1 R2 R3 Rx a b c d VSRC – + A I = 0 24 The Wheatstone Bridge  When the bridge is balanced, nodes b and c are at the same voltage and no current flows between them. The branch bc could be considered an open circuit, or a short circuit, without disturbing the rest of the circuit R1 R2 R3 Rx a b c d VSRC – + A I = 0 25 The Wheatstone Bridge If bc is open circuited, then from voltage division, and At bridge balance, Vcd = Vbd, so that That is = R1 R2 R3 Rx a b c d VSRC – + A I = 0 SRC x x cd R V R R V 2   SRC bd R V R R V 3 1 3   R2 R R x x  3 1 3 R R R  3 2 1 R R R R x  Since bc can be considered a short circuit: Vc=Vb 26 Example R1 R2 R3 Rx a b c d VSRC – + A I = 0 27 Question: Is there an electrically equivalent circuit that can simplify the analysis? 28 -Y Transformation (other names)  The -Y Transformation is also known as -T transformation. 29 30 31 Find R1, R2 and R3 in terms of Ra, Rb, Rc Rab Rab 32 33 In Short c R1 R2 a b Rc Ra Rb R3 ∆ to Y conversion Y to ∆ conversion 34 35 36 37 Example: It is required to obtain the resistance seen between terminals cd in the circuit 38 39 Solution: The lower abc  is transformed to its equivalent Y. This gives: R1=25*10/50 =5, R2=25*15/50=7.5, and R3=10*15/50=3. The 5  resistance is then added to the 9  to give 14 , and the 7.5  resistance is added to the 6.5  to give 14 . The two 14  resistors in parallel give 7 , which is added to the 3  to give 10  between terminals cd. What is the resistance seen between terminals ab? 40 41 42 Find vx 43 44 45