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EE101Lect3- Circuit Laws
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Ohm’s Law
Alternatively
n Resistors (Rk) in series
n Conductors (Gk) in parallel
• m Resistors (Rk) in p
v=Ri (units: v[V], i[A], R[Ω]
i=Gv where G=1/R [ʊ]
Rtotal= n1 Rk
Gtotal= n1 Gk
Rtotal=1/
EE101 Fall 2012 Lect 2- Kang
𝑚 1
1 (𝑅𝑘)
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Serial and Parallel Connections
EE101 Fall 2012 Lect 2- Kang
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Series Connection of Resistances
V = V1 + V2 = R1 x i + R2 x i = (R1 + R2) i
Thus Req = R1 + R2
When n resistors are in series, Req
EE101 Fall 2012 Lect 2- Kang
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Parallel Connection of Resistances
i = i1 + i2 = v/R1 + v/R2 = v (1/R1 + 1/R2)
Thus 1/Req = 1/R1 + 1/R2, or Geq = G1 + G2
Alternatively, Req = 1/ (1/R1 + 1/R2) = R1x R2/(R1 + R2)
Generally stating, when m resistances are in parallel,
Req=1/
𝑚 1
1 (𝑅𝑘)
EE101 Fall 2012 Lect 2- Kang
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How to find an Req?
EE101 Fall 2012 Lect 2- Kang
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What is the equivalent Resistance Req ?
Approach from the right end
1.Series resistance: 4 + 5 + 3 = 12
2.Parallel resistance: 4//12= 1/(1/4 + 1/12)= 4x12/(4+12)= 3
3.Series resistance: 3+3=6
4.Parallel resistance: 6//6= 1/(1/6+1/6)= 6x6/(6+6)=3
5.Series resistance: 4+3+3= 10 Ω (ans)
EE101 Fall 2012 Lect 2- Kang
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Another example of calculating Req
(2+2)//6= 4x6/(4+6)=2.4
Thus Req= 4 + 2.4 + 8= 14.4Ω
EE101 Fall 2012 Lect 2- Kang
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What is the resistance between terminals
a and b (Rab)?
EE101 Fall 2012 Lect 2- Kang
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What is Rab?
1P) 20//5= 20x5/(20+5)=4
2S) 4+1= 5
3P) 5//20= 5x20/(5+20)=4
4S) 4+2=6
5P) 9//6=9x6/(9+6)=54/15=5.6
6P) 18//5.6=18x5.6/(18+5.6)=4.27
7S) 16+4.27= 20.27Ω= Req
EE101 Fall 2012 Lect 2- Kang
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Kirchhoff’s Laws
The sum of voltages in a loop is zero.
(𝑣𝑜𝑙𝑡𝑎𝑔𝑒𝑠 𝑖𝑛 𝑎 𝑙𝑜𝑜𝑝) = 0
This is called Kirchhoff’s Voltage Law (KVL)
The Sum of the currents at a node is zero.
(what goes into a node comes out, that is
𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑛𝑤𝑎𝑟𝑑 = ∑𝑐𝑢𝑟𝑟𝑒𝑛𝑡 out𝑤𝑎𝑟𝑑
Another way of saying this is
(𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝑠 𝑖𝑛𝑡𝑜 𝑎 𝑛𝑜𝑑𝑒) = 0
This is called Kirchhoff’s Current Law (KCL)
EE101 Fall 2012 Lect 2- Kang
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What is the current through R1?
Find an equivalent resistance of R2 and R3
In parallel
R2//R3= 1/(1/R2 + 1/R3)= 1/(1/30 + 1/60)
= 20Ω
The total resistance across Vs = 10 + 20=30Ω
The current through R1 = 90V/30Ω= 3A
What about currents through R2 and R3?
3A is divided into two branchesThrough R2 is 2X of that R3, and thus 2A
through R2 and 1A through R3.
EE101 Fall 2012 Lect 2- Kang
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Finding Currents using Nodal Analysis
SOLUTION:
Problem:
N1
Let us note the voltage at node N1 as V1,
Then we have the following nodal equationBy KCL (sum of entering current s=sum of leaving currents
At each node):
(90-V1)/10 = V1/30 + V1/60
Multiplying both sides by 60, we obtain
6(90-V1)=2 V1 + V1 = 3 V1
540= 9 V1
Thus V1= 60V
[Q] Find is the current through R3
[Q] Find the power dissipated in R3
iR3 = 60/60= 1A
iR2 = 60/30= 2A
iR1 = (90-60)/10= 3A
The power dissipation in R3 is
P3= V1xiR3 = 60V x 1A= 60W.
EE101 Fall 2012 Lect 2- Kang
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∆-Y Transformation
EE101 Fall 2012 Lect 2- Kang
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∆-Y Transformation
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