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Chapter 2 Circuit analysis techniques 2023sem1
Engineering Electrical Fundamentals (ENSC2003)
University of Western Australia
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Chapter 2
Circuit analysis techniques
Basic circuit analysis using Ohm’s law
####### Evaluating unknown currents or voltages
####### Consider a generic resistive circuit connected to a power source as shown in the figures
####### below:
####### A simple approach to achieve this is to examine the circuit and represent it by an
####### effective resistance (𝑅𝑖𝑛) at the input terminals of the circuit.
####### Evaluating unknown currents or voltages
####### Consider a generic resistive circuit connected to a power source as shown in the figures
####### below:
####### A simple approach to achieve this is to examine the circuit and represent it by an
####### effective resistance (𝑅𝑖𝑛) at the input terminals of the circuit.
Start with simple circuit analysis using Ohm’s law
voltage source current source
####### Knowing the input resistance, the unknown voltage or current can be evaluated via
####### Ohm’s law,
####### 𝑉 = 𝐼𝑅𝑖𝑛
####### or
𝐼 = 𝑉 𝑅𝑖𝑛 .
####### Note also: Having an unknown resistive circuit, the effective input resistance at any
####### pair of terminals can be found by way of applying a known current (or voltage) source
####### at that terminal, measuring the resulting voltage (or current), and applying
𝑅𝑖𝑛 = 𝑉 𝐼.
####### Knowing the input resistance, the unknown voltage or current can be evaluated via
####### Ohm’s law,
####### 𝑉 = 𝐼𝑅𝑖𝑛
####### or
𝐼 = 𝑉 𝑅𝑖𝑛 .
####### Note also: Having an unknown resistive circuit, the effective input resistance at any
####### pair of terminals can be found by way of applying a known current (or voltage) source
####### at that terminal, measuring the resulting voltage (or current), and applying
𝑅𝑖𝑛 = 𝑉 𝐼.
####### Given the evaluated source voltage above, the input resistance is
####### 𝑅𝑖𝑛 =
####### 𝐼 𝑅 1 + 𝑅 2
####### 𝐼
####### = 𝑅 1 + 𝑅 2.
####### Extending this intuitively to multiple series resistors, we can state that the equivalent
####### resistance (𝑅𝑒𝑞) of a circuit containing 𝑁 resistors in series combination, is the sum of
####### the individual resistances,
####### (2)
####### Given the evaluated source voltage above, the input resistance is
####### 𝑅𝑖𝑛 =
####### 𝐼 𝑅 1 + 𝑅 2
####### 𝐼
####### = 𝑅 1 + 𝑅 2.
####### Extending this intuitively to multiple series resistors, we can state that the equivalent
####### resistance (𝑅𝑒𝑞) of a circuit containing 𝑁 resistors in series combination, is the sum of
####### the individual resistances,
####### 𝑅 (2)
𝑒𝑞 = σ𝑖= 𝑁 𝑅𝑖 = 𝑅 1 + 𝑅 2 + ⋯ + 𝑅𝑁.
Voltage division
####### Continued with the previous example. How much is the voltage across R 1 and R 2?
####### The total current driven by the the source is : 𝐼 =
𝑉 𝑅𝑖𝑛 = 𝑉 𝑅 1 +𝑅 2 .
####### Then, we have : 𝑉 1 =
𝑉 𝑅 1 +𝑅 2
####### 𝑅 1 =
𝑅 1 𝑅 1 +𝑅 2 𝑉 (2)
####### Similarly, we also have: 𝑉 2 =
𝑉 𝑅 1 +𝑅 2
####### 𝑅 2 =
𝑅 2 𝑅 1 +𝑅 2
####### 𝑉 (2)
Extension: where an applied voltage (𝑉) is divided across 𝑁 resistors in series. The
####### 𝑖
𝑡ℎ
####### resistor drops a fraction of the total voltage, given by
(2)
Voltage division
####### Continued with the previous example. How much is the voltage across R 1 and R 2?
####### The total current driven by the the source is : 𝐼 =
𝑉 𝑅𝑖𝑛 = 𝑉 𝑅 1 +𝑅 2 .
####### Then, we have : 𝑉 1 =
𝑉 𝑅 1 +𝑅 2
####### 𝑅 1 =
𝑅 1 𝑅 1 +𝑅 2 𝑉 (2)
####### Similarly, we also have: 𝑉 2 =
𝑉 𝑅 1 +𝑅 2
####### 𝑅 2 =
𝑅 2 𝑅 1 +𝑅 2
####### 𝑉 (2)
Extension: where an applied voltage (𝑉) is divided across 𝑁 resistors in series. The
####### 𝑖
𝑡ℎ
####### resistor drops a fraction of the total voltage, given by
(2) 𝑉𝑖 = 𝑅𝑖 𝑅 1 + 𝑅 2 + ⋯ + 𝑅𝑁 𝑉.
####### Given the evaluated source current above, the input resistance can be expressed as
1 𝑅𝑖𝑛 = 𝑉 ൗ 1 𝑅 1 + 1 𝑅 2 𝑉 = 1 𝑅 1 + 1 𝑅 2 .
####### For the two-resistor case, the input resistance can be written as
𝑅𝑖𝑛 = 𝑅 1 𝑅 2 𝑅 1 + 𝑅 2 .
Extension: the equivalent resistance (𝑅𝑒𝑞) of a circuit containing 𝑁 resistors in
####### parallel combination, can be expressed as
(2)
####### Given the evaluated source current above, the input resistance can be expressed as
1 𝑅𝑖𝑛 = 𝑉 ൗ 1 𝑅 1 + 1 𝑅 2 𝑉 = 1 𝑅 1 + 1 𝑅 2 .
####### For the two-resistor case, the input resistance can be written as
𝑅𝑖𝑛 = 𝑅 1 𝑅 2 𝑅 1 + 𝑅 2 .
Extension: the equivalent resistance (𝑅𝑒𝑞) of a circuit containing 𝑁 resistors in
####### parallel combination, can be expressed as
(2) 1 𝑅𝑒𝑞 = 𝑖= 𝑁 1 𝑅𝑖 = 1 𝑅 1 + 1 𝑅 2
- ⋯ + 1 𝑅𝑁 .
Current division
####### Now let us drive the same parallel resistor circuit by a
####### current source, and see the current through R 1 and R 2 :
####### The input resistance given by 𝑅𝑖𝑛 =
𝑅 1 𝑅 2 𝑅 1 +𝑅 2
####### ,
####### So, the total voltage generated by the the source is:
####### 𝑉 = 𝐼𝑅𝑖𝑛 = 𝐼
####### 𝑅 1 𝑅 2
####### 𝑅 1 + 𝑅 2
####### Then, we have: 𝐼 1 =
𝑉 𝑅 1
####### = 𝐼 ൗ
𝑅 1 𝑅 2 𝑅 1 +𝑅 2
####### 𝑅 1 =
𝑅 2 𝑅 1 +𝑅 2
####### 𝐼
####### , and 𝐼 2 =
𝑉 𝑅 2
####### = 𝐼 ൗ
𝑅 1 𝑅 2 𝑅 1 +𝑅 2
####### 𝑅 2 =
𝑅 1 𝑅 1 +𝑅 2
####### 𝐼
Current division
####### Now let us drive the same parallel resistor circuit by a
####### current source, and see the current through R 1 and R 2 :
####### The input resistance given by 𝑅𝑖𝑛 =
𝑅 1 𝑅 2 𝑅 1 +𝑅 2
####### ,
####### So, the total voltage generated by the the source is:
####### 𝑉 = 𝐼𝑅𝑖𝑛 = 𝐼
####### 𝑅 1 𝑅 2
####### 𝑅 1 + 𝑅 2
####### Then, we have: 𝐼 1 =
𝑉 𝑅 1
####### = 𝐼 ൗ
𝑅 1 𝑅 2 𝑅 1 +𝑅 2
####### 𝑅 1 =
𝑅 2 𝑅 1 +𝑅 2
####### 𝐼
####### , and 𝐼 2 =
𝑉 𝑅 2
####### = 𝐼 ൗ
𝑅 1 𝑅 2 𝑅 1 +𝑅 2
####### 𝑅 2 =
𝑅 1 𝑅 1 +𝑅 2
####### 𝐼
####### Example – Voltage division
Evaluate the voltage across resistor 𝑅 2 :
####### Example – Voltage division
Evaluate the voltage across resistor 𝑅 2 :
####### Example – Current division
Evaluate the current through resistor 𝑅 2 :
####### Example – Current division
Evaluate the current through resistor 𝑅 2 : 12 A 𝑅 2 = 5 Ω 10 Ω 10 V 8 Ω 𝑅 2 = 12 Ω
Analysis of more complicated circuits
It is then also evident that 𝑅 1 and 𝑅 2 are in parallel, as are the combined resistances 𝑅 5 + 𝑅 6 and 𝑅 7 + 𝑅 8 + 𝑅 9. We will denote these parallel combinations as 𝑅 1 ∥ 𝑅 1 , and (𝑅 5 + 𝑅 6 ) ∥ (𝑅 7 + 𝑅 8 + 𝑅 9 ). It is then also evident that 𝑅 1 and 𝑅 2 are in parallel, as are the combined resistances 𝑅 5 + 𝑅 6 and 𝑅 7 + 𝑅 8 + 𝑅 9. We will denote these parallel combinations as 𝑅 1 ∥ 𝑅 1 , and (𝑅 5 + 𝑅 6 ) ∥ (𝑅 7 + 𝑅 8 + 𝑅 9 ). Notation: Two resistors in parallel, say 𝑅𝑎 and 𝑅𝑏, are often denoted as 𝑅𝑎 ∥ 𝑅𝑏 (or sometimes simply as 𝑅𝑎//𝑅𝑏). Algebraically, we can say 1 𝑅𝑎 ∥ 𝑅𝑏 = 1 𝑅𝑎 + 1 𝑅𝑏 .
The next simplification of this circuit is then:The next simplification of this circuit is then:
####### The final expression for input resistance of this
####### parallel combination:
####### 𝑅 1 ∥ 𝑅 2 ∥ 𝑅 3 + 𝑅 4 + 𝑅 5 + 𝑅 6 ∥ 𝑅 7 + 𝑅 8 + 𝑅 9
The next level of complexity
####### The following example presents no segmented layout.
####### The solution R in to this circuit would get very difficult if using only the tools examined
####### so far. Furthermore, this circuit still contains only a voltage source and resistors.
####### Adding other types of circuit element will further complicate analysis.
The next level of complexity
####### The following example presents no segmented layout.
####### The solution R in to this circuit would get very difficult if using only the tools examined
####### so far. Furthermore, this circuit still contains only a voltage source and resistors.
####### Adding other types of circuit element will further complicate analysis.
We need advanced circuit analysis tools / techniques!
Kirchhoff’s Laws
(Current-based and
voltage-based analyses)
Chapter 2 Circuit analysis techniques 2023sem1
Course: Engineering Electrical Fundamentals (ENSC2003)
University: University of Western Australia
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