Simple Resistive Circuits Qi Xuan Zhejiang University of Technology September 2015 Electric Circuits 1
Structure Resistors in Series Resistors in Parallel The Voltage/Current- Divider Circuit Voltage/Current Division Measuring Voltage and Current Measuring Resistance- The Wheatstone Bridge Delta- to- Wye (Pi- to- Tee) Equivalent Circuit Electric Circuits 2
A Rear Window Defroster A Rear Window Defroster A resiseve circuit Electric Circuits 3
Resistors in Series Electric Circuits 4
In general, if k resistors are connected in series, the equivalent single resistor has a resistance equal to the sum of the k resistances, or Electric Circuits 5
Resistors in Parallel Electric Circuits 6
Conductance Electric Circuits 7
Typical Cases If k = 2, we have If R 1 =R 2 = = R k = R, we have 1/R eq = k/r R eq = R/k Electric Circuits 8
Example #1 For the circuit shown, find (a) the voltage v, (b) the power delivered to the circuit by the current source, and (c) the power dissipated in the 10 Ω resistor. Electric Circuits 9
i 1 i 2 R 1 v 1 R 1 = (10+6) 64 / (10+6+64) = 12.8 Ω R = (7.2+12.8) 30 / (7.2+12.8+30) = 12 Ω v = 5R = 60 V i 1 = v / (7.2+R 1 ) = 60 / (7.2+12.8) = 3 A v 1 = i 1 R 1 = 3 12.8 = 38.4 V i 2 = v 1 / (6+10) = 38.4 / 16 = 2.4 A p = i 22 10 = 2.4 2.4 10 = 57.6 W Solu5on for Example #1 Electric Circuits 10
The Voltage- Divider Circuits Electric Circuits 11
The current- Divider Circuit Electric Circuits 12
Example #2 a) Find the value of R that will cause 4 A of current to flow through the 80 Ω resistor in the circuit shown. b) How much power will the resistor R from part (a) need to dissipate? c) How much power will the current source generate for the value of R from part (a)? Electric Circuits 13
Solu5on for Example #2 i v R i = 20R / (40+80+R) = 4 A For (a), we have R = 30 Ω For (b), we have v R = (40+80)i = 120 4 = 480 V For (c), we have v = 60 20+v R = 1200+480 = 1680 V p R = v R 2 / R= 480 2 / 30 = 7680 W p = 20v = 20 1680 = 33600 W Electric Circuits 14
Voltage/Current Division Electric Circuits 15
Example #3 a) Use voltage division to determine the voltage v 0 across the 40 Ω resistor in the circuit shown. b) Use v 0 from part (a) to determine the cur- rent through the 40 Ω resistor, and use this current and current division to calculate the current in the 30 Ω resistor. c) How much power is absorbed by the 50 Ω resistor? Electric Circuits 16
Solu5on for Example #3 i i 1 i 2 For (a), we have R eq = 40+(20 30 (50+10))+70 = 40+10+70=120 Ω v o = 60 40 / R eq = 20 V For (b), we have i = v o / 40 = 20 / 40 = 0.5 A R p For (c), we have i 2 = i R p / 60 = 0.5 10 / 60 A = 1/12 A i 1 = i R p / 30 = 0.5 10 / 30 A = 166.67 ma p 2 = 50i 22 = 50 / 144 W = 347.22 mw Electric Circuits 17
Measuring Voltage and Current An ammeter is an instrument designed to measure current; A voltmeter is an instrument designed to measure voltage An ideal ammeter has an equivalent resistance of 0 Ω and funceons as a short circuit in series with the element whose current is being measured. An ideal voltmeter has an infinite equivalent resistance and thus funceons as an open circuit in parallel with the element whose voltage is being measured. Electric Circuits 18
Digital and Analog Meters Electric Circuits 19
d Arsonval Meter Movement d'arsonval meter movement consists of a movable coil placed in the field of a permanent magnet. When current flows in the coil, it creates a torque on the coil, causing it to rotate and move a pointer across a calibrated scale. By design, the deflec5on of the pointer is directly propor5onal to the current in the movable coil. The coil is characterized by both a voltage raeng and a current raeng. For example, one commercially available meter movement is rated at 50 mv and 1 ma. This means that when the coil is carrying 1 ma, the voltage drop across the coil is 50 mv and the pointer is deflected to its full-scale position. Electric Circuits 20
DC Ammeter/Voltage Circuit Electric Circuits 21
Example #4 a) A 50 mv, 1 ma d'arsonval movement (le^) is to be used in an ammeter with a full- scale reading of 10 ma. Determine R A. b) Find the current in the circuit shown (right). c) If the ammeter (le^) is used to measure the current in the circuit (right), what will it read? Electric Circuits 22
Solu5on for Example #4 i v A i B i A For (a) (left), we have i A = I - i B =10 ma -1mA = 9 ma R A = v A / i A = 50 / 9 =5.56 Ω R eq For (b) (right), we have i = 1 / 100 A = 10 ma For (c), we have R eq = v A / i = 50 / 10 = 5 Ω i = 1 / (100+R eq ) R eq = 1 / 105 A= 9.524 ma Electric Circuits 23
Measuring Resistance: The Wheatstone Bridge Ba`ery i g = 0 i 1 = i 3 i 2 = i x i 1 R 1 = i 2 R 2 i 3 R 3 = i x R x In a commercial Wheatstone bridge, R 1 and R 2 consist of decimal values of resistances that can be switched into the bridge circuit. R 1 / R 2 = i 2 / i 1 R 3 / R x = i x / i 3 = i 2 / i 1 R x = R 2 R 3 / R 1 Electric Circuits 24
Example #5 The bridge circuit shown is balanced when R 1 =100 Ω, R 2 =1000 Ω, and R 3 =150 Ω. The bridge is energized from a 5 V dc source. a) What is the value of R x? b) Suppose each bridge resistor is capable of dissipaeng 250 mw. Can the bridge be le^ in the balanced state without exceeding the power- dissipaeng capacity of the resistors, thereby damaging the bridge? Electric Circuits 25
Solu5on for Example #5 i 1 i 2 p 1 =i 12 R 1 = 0.02 2 100 =0.04 W p 2 =i 22 R 2 = 0.002 2 1000 =0.004 W p 3 =i 12 R 3 = 0.02 2 150 =0.06 W p x =i 22 R x = 0.002 2 1500 =0.006 W For (a), we have R x = R 2 R 3 / R 1 = 1000 150 / 100 = 1500 For (b), we have i 1 = v / (R 1 +R 3 )= 5 / 250 = 0.02 A i 2 = v / (R 2 +R x )= 5 / 2500 = 0.002 A Electric Circuits 26
Delta- to- Wye (Pi- to- Tee) Equivalent Circuits Δ π Y Electric Circuits 27 T
The Δ- to- Y Transforma5on R 1 = (R ab +R ca -R bc ) / 2 = R b R c / (R a +R b +R c ) R 2 = (R ab +R bc -R ca ) / 2 = R c R a / (R a +R b +R c ) R 3 = (R bc +R ca -R ab ) / 2 = R a R b / (R a +R b +R c ) How about Y- to- Δ? Electric Circuits 28
The Y- to- Δ Transforma5on Electric Circuits 29
Example #6 Use a Y- to- Δ transformaeon to find the voltage v in the circuit shown. Electric Circuits 30
Solu5on for Example #6 c a c R b R a b a R c b R a = (20 10+20 5+10 5) / 20 =17.5 Ω R b = (20 10+20 5+10 5) / 5 = 70 Ω R c = (20 10+20 5+10 5) / 10 = 35 Ω Electric Circuits 31
a 28Ω 70Ω R eq = 35 [(28 70)+105 17.5] = 35 (20+15) = 17.5 Ω 2A v 35Ω c 105Ω 17.5Ω v = 2R eq = 35 V b
Summary Series/Parallel resistors Voltage/Current division Digital/Analog voltmeter/ammeter Wheatstone bridge Δ- to- Y/Y- to- Δ transformaeon Electric Circuits 33