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ECE 3301 General Electrical Engineering
Presentation 03 Simple Resistive Circuits
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Resistivity Length l (m) I π
=π π π΄ Resistivity Ο (β¦-m)
Cross-sectional Area A (m2 ) Assuming the Current is uniformly distributed across the Cross-Sectional Area.
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Simple Resistive Circuits
The simplest circuit is a voltage source connected across a resistance as demonstrated below. VS IS R VR= VS
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Simple Resistive Circuits
The voltage source VS is placed across the resistance R and the current IS flows in the closed loop. VS IS R VR= VS
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Simple Resistive Circuits
Assuming VS and R are known, the current is given by Ohmβs Law. + π π β π π
=0 π π β πΌ π π
=0 πΌ π π
= π π πΌ π = π π π
VS IS R VR= VS
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Resistances in Series VS IS V1 V2 R1 R2 I1 I2 A slightly more complicated resistive circuit has two resistors in series.
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Resistances in Series Note that resistors R1 and R2 are connected end-to-end. In this configuration they share the same current IS = I1 = I2 VS IS V1 V2 R1 R2 I1 I2
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Series Connection When the same continuous current flows through circuit elements, those circuit elements are connected in series.
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Resistances in Series VS IS V1 V2 R1 R2 The voltage source VS is placed across the series combination of R1 and R2. The current IS is established in the closed loop.
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Resistances in Series VS IS V1 V2 R1 R2 There will be a voltage V1 βdroppedβ across resistor R1 There will be a voltage V2 βdroppedβ across resistor R2
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Resistances in Series VS IS V1 V2 R1 R2 Applying Kirchhoffβs Voltage Law to this closed loop gives the equation + π π β π 1 β π 2 =0 π π = π 1 + π 2
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Resistances in Series VS IS V1 V2 R1 R2 Using Ohmβs Law, we may express the voltage across each resistance by the product of the current and resistance, IR. This gives the equation π π = πΌ π π
1 + πΌ π π
2
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Resistances in Series Factoring IS reveals π π = πΌ π π
1 + π
2
VS IS V1 V2 R1 R2 Factoring IS reveals π π = πΌ π π
1 + π
2 The quantity in parentheses represents the equivalent resistance of the series connection of R1 and R2.
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Resistances in Series π
eq = π
1 + π
2
The current in the loop is given by πΌ π = π π π
eq VS Req
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The same current, I, flows through the resistors.
Vab I V1 V2 R1 R2 Vk Rk a b This same procedure may be applied to any number of resistors connected in series. The same current, I, flows through the resistors. The total voltage from point a to point b is Vab.
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Vab I V1 V2 R1 R2 Vk Rk a b Applying Kirchhoffβs Voltage Law to this combination of resistors leads to the equation: π ab = π 1 + π 2 + βββ+ π k
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Using Ohmβs Law, the same voltage may be expressed
Vab I V1 V2 R1 R2 Vk Rk a b Using Ohmβs Law, the same voltage may be expressed π ab =πΌ π
1 +πΌ π
2 + βββ+πΌ π
k
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Factoring I from each term in the summation reveals
Vab I V1 V2 R1 R2 Vk Rk a b Factoring I from each term in the summation reveals π ab =πΌ π
1 + π
2 + βββ+ π
k The term in parentheses represents the equivalent resistance of the series combination of resistances.
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I a π
eq = π
1 + π
2 + βββ+ π
k π
eq = π=1 π π
π Vab Req b
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Resistances in Series For resistors in series, the equivalent resistance is equal to the sum of the resistances of the individual resistors. For resistances in series, the equivalent resistance is always larger than the largest resistance in the group.
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Resistances in Parallel
Consider the resistive circuit below: VS IS V1 R1 V2 R2 I1 I2 Node
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Resistances in Parallel
Note that resistances R1 and R2 are connected side-by side. The voltage source VS is placed across both R1 and R2. VS IS V1 R1 V2 R2 I1 I2 Node
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Resistances in Parallel
+ π π β π 1 =0 π π = π 1 + π π β π 2 =0 π π = π 2 + π 1 β π 2 =0 π 1 = π 2 VS IS V1 R1 V2 R2 I1 I2 Node π π = π 1 = π 2
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Parallel Connection When the same voltage appears across circuit elements, those circuit elements are connected in parallel.
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Resistances in Parallel
Applying Kirchhoffβs Current Law to the node results in the equation: πΌ π β πΌ 1 β πΌ 2 =0 πΌ π = πΌ 1 + πΌ 2 VS IS R1 R2 I1 I2
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Resistances in Parallel
By applying Ohmβs Law, we can replace the current in each resistance by the quotient of the voltage divided by the resistance, V/R. VS IS R1 R2 I1 I2
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Resistances in Parallel
πΌ S = π S π
π S π
2 VS IS R1 R2 I1 I2
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Resistances in Parallel
The term VS can be factored out of both terms. πΌ S = π S 1 π
π
2 VS IS R1 R2 I1 I2
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Resistances in Parallel
The term in brackets has the units of conductance, the reciprocal of resistance. This suggests an equivalent conductance: πΊ eq = 1 π
eq = 1 π
π
2 πΊ eq = πΊ 1 + πΊ 2
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Resistances in Parallel
Putting the term in brackets over a common denominator leads to the equation: 1 π
eq = π
2 + π
1 π
1 π
2 π
eq = π
1 π
2 π
1 + π
2
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Resistances in Parallel
This connection occurs frequently, so it is worthwhile to remember that for two resistors in parallel : πΈππ’ππ£πππππ‘ π
ππ ππ π‘ππππ= πππππ’ππ‘ ππ π‘βπ π
ππ ππ π‘πππππ ππ’π ππ π‘βπ π
ππ ππ π‘πππππ
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Resistances in Parallel
An arbitrary number of resistances connected in parallel. Vab I R1 R2 I1 I2 Rk Ik βββ Node
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Resistances in Parallel
The same voltage Vab appears across each of the resistances. Vab I R1 R2 I1 I2 Rk Ik βββ Node
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Resistances in Parallel
Applying Kirchhoffβs current Law at the node results in the equation: πΌ= πΌ 1 + πΌ 2 + βββ+ πΌ π Vab I R1 R2 I1 I2 Rk Ik βββ Node
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Resistances in Parallel
Applying Ohmβs Law, we can replace the current through each resistance with the quotient of the voltage divided by the resistance, V/R. Vab I R1 R2 I1 I2 Rk Ik βββ
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Resistances in Parallel
πΌ= π ab π
π ab π
2 + βββ+ π ab π
π Factoring Vab from each term leads to: πΌ= π ππ 1 π
π
2 + βββ+ 1 π
π
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Resistances in Parallel
πΌ= π ππ 1 π
π
2 + βββ+ 1 π
π The term in brackets has the units of conductance and can be expressed πΊ eq = 1 π
eq = 1 π
π
2 + βββ+ 1 π
π
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Resistances in Parallel
1 π
eq = 1 π
π
2 + βββ+ 1 π
π Vab I R1 R2 I1 I2 Rk Ik βββ
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Resistances in Parallel
1 π
eq = 1 π
π
2 + βββ+ 1 π
π 1 π
eq = π=1 π 1 π
π πΊ eq = π=1 π πΊ π Vab I Req
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Resistances in Parallel
For resistors in parallel, the equivalent conductance is equal to the sum of the conductances of the individual resistors. For resistances in parallel, the equivalent resistance is always smaller than the smallest resistance in the group.
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Resistances in Parallel
Warning: Do not make these common mistakes! π
eq β 1 π
π
2 + βββ+ 1 π
π π
eq β π
1 π
2 π
3 π
1 + π
2 + π
3 These are wrong!
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Simplifying Resistive Circuits
VS R2 R4 Are R1 and R3 connected in series? No! They do not share the same current!
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Simplifying Resistive Circuits
VS R2 R4 Are R2 and R4 connected in parallel? No! They do not share the same voltage!
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Simplifying Resistive Circuits
VS R2 R4 Are R1 and R2 connected in series? Β No! They do not share the same current! Β Are R1 and R2 connected in parallel? No! They do not share the same voltage!
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Simplifying Resistive Circuits
VS R2 R4 Are R3 and R4 connected in series? Yes! They share the same current!
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Simplifying Resistive Circuits
VS R2 R3 + R4
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Simplifying Resistive Circuits
R2 || (R3 + R4) VS
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Simplifying Resistive Circuits
R1 + R2 || (R3 + R4) VS π
eq = π
1 + π
2 π
3 + π
4 π
2 + π
3 + π
4 π
eq = π
1 π
2 + π
3 + π
4 + π
2 π
3 + π
4 π
2 + π
3 + π
4
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Simplifying Resistive Circuits
VS R1 R2 R3 R4 R5
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Simplifying Resistive Circuits
VS R2 + R3 R1 R4
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Simplifying Resistive Circuits
VS R5 || (R2 + R3) R1 R4
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Simplifying Resistive Circuits
VS R4 + R5 || (R2 + R3) R1
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Simplifying Resistive Circuits
VS R1||{R4 + R5 || (R2 + R3)} π
eq = π
1 π
4 + π
5 π
2 + π
3 π
5 + π
2 + π
π
1 + π
4 + π
5 π
2 + π
3 π
5 + π
2 + π
3
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Series or Parallel? R1 R2 R3 R4 R5 R6 R7 R8 VS
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Series or Parallel? R2 R3 R1 R4 R7 R8 R5 VS R6
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