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Lecture (02)Basic laws II ‐ Circuit
SimplificationBy:
Dr. Ahmed ElShafee
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١
Agenda• Series Resistors and Voltage Division• Parallel Resistors and Current Division• Wye‐Delta Transformations• Applications
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢
Series Resistors and Voltage Division• The process of combining the resistors is done by combining
two of them at a time.
• the same current i flows in both of them. • Applying Ohm’s law to each of the resistors
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣
• apply KVL to the loop
• Combining Eqs.
• can be written as
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤
• The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances.
• Voltage deviation
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥
• In general, if a voltage divider has N resistors in series with the source voltage v, the nth resistor ( Rn) will have a voltage drop of
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٦
Parallel Resistors and Current Division• Ohm’s law
• Applying KCL at node a
• Substituting Eq.
• where Req is the equivalent resistance of the resistors in parallel
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٧
• The equivalent resistance of two parallel resistors is equal to the product of their resistances divided by their sum.
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٨
Example 01•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٩
• 6Ω || 3Ω = 2Ω
• 1Ω ‐‐ 5Ω = 6Ω
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١٠
• 2Ω ‐‐ 2Ω = 4Ω
• 4Ω || 6Ω = 2.4Ω
• 4Ω ‐‐ 2.4Ω ‐‐ 8Ω = 14.4Ω
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١١
•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١٢
• 4Ω ‐‐ 5Ω ‐‐ 3Ω = 12Ω
• 12Ω || 4Ω = 3Ω
• 3Ω ‐‐ 3Ω = 6Ω
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١٣
• 6Ω || 6Ω = 3Ω
• 4 Ω ‐‐ 3 Ω ‐‐ 3 Ω = 10Ω
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١٤
Example 03• Find Req
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١٥
• 1Ω ‐‐ 5Ω = 6 Ω
• 6Ω || 4Ω = 2.4 Ω
• 2.4Ω || 12Ω = 2Ω
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١٦
• (1 Ω ‐‐ 2 Ω) || 6 Ω = 2 Ω
• (2 Ω || 3 Ω) ‐‐ 10 Ω = 11.2 Ω
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١٧
Example 04• Find Req
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١٨
• 20 || 5 = 4
• (4 ‐‐ 1) || 20 = 4
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits١٩
• 4 ‐‐ 2 = 6
• 6 || 9 || 18 = 3.6 || 18 = 3
• 3 ‐‐ 16 = 19
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢٠
Example 05
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢١
• 3 || 6 = 2
• Voltage division :V2Ω = 12 x 2/(2+4) = 4 V
• Return backV2Ω = V6Ω = V3Ω = 4 V
• OhmI3Ω = 4/3 = 1.333 AmpP3Ω = 4 x 1.333 = 5.33 W
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢٢
Example 06
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢٣
• 12 Ω || 6 Ω = 4 Ω• 10 Ω || 40 Ω = 8 Ω
• Voltage division :V8Ω = 30 x 8 /(8+4) = 20 VV4Ω = 30 x 4 /(4+8) = 10 V
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢٤
• Returning step back:V4Ω = V12Ω = V6Ω = 10 VoltI12Ω= 10/12 = 0.833 AmpP12Ω = 8.33 W
V8Ω = V10Ω = V40Ω = 20 VoltI40Ω = 20/40 = 0.5 AmpW40Ω = 10 W
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢٥
Example 07•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢٦
• 12 kΩ ‐‐ 6 kΩ = 18 kΩ
• 18 kΩ || 9 kΩ = 6 kΩ
• Current division:I9kΩ = 30 x 18 (9+18) = 20 mAI18kΩ = 30 x 9 (9+18) = 10 mA
• OhmV9kΩ = 9 x1000x 20 = 180 VP9kΩ = (20 x 10‐3)2 x 9x1000 = 3.6 W
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢٧ Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢٨
• I18kΩ = I6kΩ = I12kΩ = 30 x 9 (9+18) = 10 mA
• P6kΩ = (10 x 10‐3)2 x 6 x 1000 = 0.6 W
• P12kΩ = (10 x 10‐3)2 x 12 x 1000 = 1.2 W
Example 08
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٢٩ Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣٠
• 1 kΩ ‐‐ 3 kΩ = 4 k Ω
• 20 kΩ || 5 kΩ = 4 kΩ
• Current divisionI4kΩ = 30 /2 = 15 mA
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣١
• One step back• Voltage divisionV3kΩ = 3 x 1000 x 15 / 1000 = 45 Volt
• Back one step• Current division:I20KΩ = 15 x 5/(5+20) = 3 mA• ohmV20KΩ = 3 x 10‐3 x 20 x 1000 = 60 Volt
• P20KΩ = 3 x 60 = 180 mW• P3KΩ = 15 x 45 = 675 mW
• P30mA = V30mA x I30mA = 30 x V20KΩ = 30 x 60 = 180 mW
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣٢
Wye‐Delta Transformations
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣٣
• Situations often arise in circuit analysis when the resistors are neither in parallel nor in series like following bridge
• Many circuits of the type can be simplified by using three‐terminal equivalent networks Wye‐Delta transformation
• They are used in three‐phase networks, electrical filters, and matching networks.
•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣٤
1 2
3
a
b
c
•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣٥
•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣٦
Each resistor in the Y network is the product of the resistors in the twoadjacent Δ branches, divided by the sum of the three Δ resistors.
•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣٧
Each resistor Δ in the network is the sum of all possible products of Yresistors taken two at a time, divided by the opposite Y resistor.
Example 09•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣٨
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٣٩
Example 10
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤٠
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤١
Example 11•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤٢
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤٣
30 || 70 = 21
17.5 || 12.5 = 7.3
15 || 35 = 10.5
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤٤
10.5 ‐‐ 7.3 = 17.8
17.8 || 21 = 9.63
Ohm:I = 120/9.63 = 12.46 amp
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤٥
Example 12•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤٦
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤٧
60 || 10 = 8.5775 || 50 = 30
8.57 ‐‐ 30 = 38.57
38.57 || 90 = 26.99
13‐‐ 27= 40Ohm:I = 240/40 = 6 Amp
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤٨
Applications : Lighting system• Lighting systems such as in a house often consist of N lamps
connected either in parallel or in series
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٤٩
• The series connection is easy to manufacture but is not used in practice, for at least two reasons. – First, it is less reliable; when a lamp fails, all the lamps go out.
– Second, it is harder to maintain; when a lamp is bad, one must test all the
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥٠
Example 13•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥١ Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥٢
• V90W = 9 V• I90W = P/V = 20/9 = 2.22 Amp• R90W = V/I = 9/2.22 = 4 Ohm
• P15W,10W = I(V15w + V10W) = V15W x I + V10W x I = P15W + P10W• I15W = I10W= (P15W + P10W) / (V15W+V10W) = 25/(9)=2.78 Amp• R15W = P/I2 = 15/7.73 = 1.94 Ohm• R10W = P/I2 = 10/7.73 = 1.23 Ohm
•
Example 14•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥٣
• Vlamp = 110V• Ilamp = P/V = 40/110 = 0.36 Amp
• Vlamp = 110/10 = 11 volt• Ilamp = 40/11 = 3.6 Amp
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥٤
Applications : Design of DC Meters• Digital meters:
– Digital meters, get samples of voltage or current, time separation between sample is very small.
– Converting these samples from analog form to digital form, in order to display values on LCD.
– Digital meters are very sensitive and have a very high accuracy and tolerance.
– It has a minor (neglected) effect on voltageor current of the circuit under test.
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥٥ Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥٦
• Analog meters:– Analog meters based on basic devicecalled D’arsonval meter.– Which is a movable coil placed in permanent magnetic field.– When current passes through the coil, coil deflect.
– Deflection directly proportional to voltage rating and current rating.
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥٧
• We need to determine the value of Rn to be connected in series with the internal resistance Rm of the voltmeter. • In any design, we consider the worst‐case condition Ifs = Im• In this case, the worst case occurs when the full‐scale current
flows through the meter.• This should also correspond to the maximum voltage reading
or the full‐scale voltage • Since the multiplier resistance is in series with the internal
resistance
•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥٨
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٥٩
• Similarly, the ammeter measures the current through the load and is connected in series with it• We notice that Rm and Rn are in parallel and that at full‐scale
reading
•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٦٠
• Rx The resistance of a linear resistor can be measured in two ways.
• An indirect way is to measure the current I that flows through it by connecting an ammeter in series with it and the voltage V across it by connecting a voltmeter in parallel with it,
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٦١ Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٦٢
• An ohmmeter consists basically of a d’Arsonval movement, a variable resistor or potentiometer, and a battery
• The resistor R is selected such that the meter gives a full‐scale deflection Im = Ifs ; that is, when Rx=0.
• Substituting
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٦٣
Example 15•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٦٤
•
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٦٥
Example 16
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٦٦
a) Ifs = 1 AIm = 1 mARn = (1/(1‐0.001) 50) = 50.05 mOhm
b) Ifs = 100mARn = (1/(0.1‐0.001) 50) = 505.05 mOhm
c) Ifs = 10 mARn = (1/(0.01‐0.001) 50) = 5555.56 mOhm
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٦٧
Thanks,..See you next week (ISA),…
Dr. Ahmed ElShafee, ACU : Spring 2016, Electric Circuits٦٨
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