DE4401 Electrical and Electronics Principles 1
APTE 5601 Electrical and Electronics Principles
Lab 3 Answers
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Question 1
Analyse the following circuit using:
a) Branch current method
b) Loop current method
c) Multisim simulation software.
Solution: see the next page...
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DE4401 Lab3 Answers
Solution:
a) Branch current method:
1. Choose a node and assume directions of currents.
2. Write a KCL equation relating currents at the node:
𝐼1 + 𝐼2 − 𝐼3 = 0
3. Label resistor voltage drop polarities based on assumed currents.
4. Write KVL equations for each loop of the circuit, substituting the product IR for E in
each resistor term of the equations.
𝐸1 − 𝑅1 𝐼1 − 𝐸2 = 0
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DE4401 Lab3 Answers
𝐸2 − 𝑅2 𝐼3 = 0
5. Solve for unknown branch currents (simultaneous equations).
Three equations from which we find three unknown currents are:
equation(1): 𝐼1 + 𝐼2 − 𝐼3 = 0
equation(2): 𝐸1 − 𝑅1 𝐼1 − 𝐸2 = 0
equation(3): 𝐸2 − 𝑅2 𝐼3 = 0
First we enter the values we know (Rs and Es from the schematic):
equation(1): 𝐼1 + 𝐼2 − 𝐼3 = 0
equation(2):
85 − 10𝐼1 − 45 = 0
equation(3): 45 − 5𝐼3 = 0
From the equation (2) we can calculate current I1 directly as:
85 − 45 40
𝐼1 =
=
= 4𝐴
10
10
Thus:
𝐼1 = 4𝐴
From the equation (3) we can calculate current I3 directly as:
𝐸2 45
𝐼3 =
=
= 9𝐴
𝑅2
5
Thus:
𝐼3 = 9𝐴
From the equation (1) we calculate current I2
𝐼2 = 𝐼3 − 𝐼1 = 9 − 4 = 5𝐴
So, 𝐼2 = 5𝐴
If any solution is negative, then the assumed direction of current for that solution is wrong:
this is not the case here, all assumed direction are the actual current directions.
6. Solve for voltage drops across all resistors (V=IR).
Using Ohm’s law:
𝑉1 = 𝑅1 𝐼1
𝑉1 = 10 ∙ 4 = 40V
𝑉2 = 𝑅2 𝐼3
𝑉2 = 5 ∙ 5 = 25V
Conclusion (how to the simulation compare with calculation):
………………………………………………………………………………………………..
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DE4401 Lab3 Answers
a) Loop current method
Steps to follow for the “Loop (Mesh) Current” method of network analysis:
1. Draw currents in loops of circuit, to account for all components.
2. Label resistor voltage drop polarities based on assumed directions of loop currents.
3. Write KVL equations for each loop, substituting the product IR for E in each resistor
term of the equation. Where two mesh currents intersect through a component,
express the current as the algebraic sum of those two mesh currents (i.e. I1 + I2 if the
currents go in the same direction through that component or I 1 - I2 if not.)
Two equations from which we find two unknown currents are:
equation(1): 𝐸1 − 𝑅1 𝐼1 − 𝐸2 = 0
equation(2):
𝐸2 − 𝑅2 𝐼2 = 0
4. Solve for unknown loop currents (simultaneous equations). If any solution is negative,
the assumed current direction is wrong.
get the numbers in:
equation(1): 85 − 10𝐼1 − 45 = 0
equation(2): 45 − 5𝐼2 = 0
add the likes together, sort out the equations (the “knowns” on one side, the “unknowns” on
the other side of “=”:
equation(1): 10𝐼1 = 40
equation(2): 5𝐼2 = 45
And solve : 𝐼1 = 4𝐴 and
𝐼2 = 9𝐴
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DE4401 Lab3 Answers
5. Algebraically add loop currents to find current in components which share multiple
loop currents.
The only unknown is current in the branch containing E2 source. That current is obtained
from two loop currents going through that branch:
𝐼3 = 𝐼2 − 𝐼1 = 9 − 4 = 5𝐴
The current in this branch is in direction of loop current I2.
6. Solve for voltage drops across all resistors (V=IR).
Using Ohm’s law:
𝑉1 = 𝑅1 𝐼1
𝑉1 = 10 ∙ 4 = 40V
𝑉2 = 𝑅2 𝐼3
𝑉2 = 5 ∙ 5 = 25V
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DE4401 Lab3 Answers
Question 2
Analyse the following circuit using:
a) Branch current method
b) Loop current method
c) Multisim simulation software.
Solution: go to the next page
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DE4401 Lab3 Answers
Branch current method:
Steps to follow for the “Branch Current” method of network analysis:
1. Choose a node and assume directions of currents.
2. Write a KCL equation relating currents at the node.
equation(1):
𝐼1 + 𝐼2 − 𝐼3 = 0
3. Label resistor voltage drop polarities based on assumed currents.
4. Write KVL equations for each loop of the circuit, substituting the product IR for E in
each resistor term of the equations.
equation(2): 𝐸1 − 𝑅1 𝐼1 − 𝐸2 + 𝑅2 𝐼2 = 0
equation(3): 𝐸2 − 𝑅2 𝐼2 − 𝑅3 𝐼3 − 𝑅4 𝐼3 = 0
5. Solve for unknown branch currents (simultaneous equations).
equation(1): 𝐼1 + 𝐼2 − 𝐼3 = 0
equation(2): 𝐸1 − 𝑅1 𝐼1 − 𝐸2 + 𝑅2 𝐼2 = 0
equation(3): 𝐸2 − 𝑅2 𝐼2 − 𝑅3 𝐼3 − 𝑅4 𝐼3 = 0
get the numbers in:
equation(1): 𝐼1 + 𝐼2 − 𝐼3 = 0
equation(2): 110 − 5𝐼1 − 190 + 5𝐼2 = 0
equation(3): 190 − 5𝐼2 − 15𝐼3 − 20𝐼3 = 0
sort out the “likes”:
equation(1):
equation(2):
equation(3):
𝐼1 + 𝐼2 − 𝐼3 = 0
−5𝐼1 + 5𝐼2 = 80
5𝐼2 + 35𝐼3 = 190
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DE4401 Lab3 Answers
6. If any solution is negative, then the assumed direction of current for that solution is
wrong.
7. Solve for voltage drops across all resistors (V=IR).
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DE4401 Lab3 Answers
Loop current method
Steps to follow for the “Loop (Mesh) Current” method of network analysis:
1. Draw currents in loops of circuit, to account for all components.
2. Label resistor voltage drop polarities based on assumed directions of loop currents.
3. Write KVL equations for each loop, substituting the product IR for E in each resistor
term of the equation. Where two mesh currents intersect through a component, express
the current as the algebraic sum of those two mesh currents (i.e. I1 + I2 if the currents
go in the same direction through that component or I 1 - I2 if not.)
equation(1):
equation(2):
𝐸1 − 𝑅1 𝐼1 − 𝐸2 − 𝑅2 (𝐼1 + 𝐼2 ) = 0
𝑅4 𝐼2 + 𝑅3 𝐼2 + 𝐸2 + 𝑅2 (𝐼1 + 𝐼2 ) = 0
4. Solve for unknown loop currents (simultaneous equations). If any solution is negative,
the assumed current direction is wrong.
put the numbers in:
equation(1):
equation(2):
110 − 5𝐼1 − 190 − 5(𝐼1 + 𝐼2 ) = 0
20𝐼2 + 15𝐼2 + 190 + 5(𝐼1 + 𝐼2 ) = 0
add the like terms:
equation(1):
equation(2):
10𝐼1 + 5𝐼2 = −80 /∙ (−8)
5𝐼1 + 40𝐼2 = −190
multiplying equation (1) by (-8)
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DE4401 Lab3 Answers
−80𝐼1 − 40𝐼2 = 640
5𝐼1 + 40𝐼2 = −190
equation(1):
equation(2):
adding two equations together eliminates I2
−75𝐼1 = 450
𝐼1 = −6𝐴
(assumed current direction is wrong)
Return to the equation (1) and include the value of current I1 to calculate current I2:
10𝐼1 + 5𝐼2 = −80
−60 + 5𝐼2 = −80
5𝐼2 = −20
𝐼2 = −4𝐴
(assumed current direction is wrong)
5. Algebraically add loop currents to find branch current in components which share
multiple loop currents.
𝐼b3 = 𝐼1 + 𝐼2
𝐼b3 = −6 − 4
𝐼b3 = −10𝐴
We can also see that the remaining branch currents are equal to the loop currents:
𝐼b1 = |𝐼1 |
and
𝐼b2 = |𝐼2 |
The branch currents and the actual direction are given in the Figure below:
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DE4401 Lab3 Answers
Question 3
Analyse the following circuit using Loop current method. Use Multisim simulation
software to confirm your calculations.
Solution for loop current method:
1. Draw currents in loops of circuit, to account for all components.
2. Label resistor voltage drop polarities based on assumed directions of loop currents.
3. Write KVL equations for each loop, substituting the product IR for E in each resistor
term of the equation. Where two mesh currents intersect through a component, express
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DE4401 Lab3 Answers
the current as the algebraic sum of those two mesh currents (i.e. I1 + I2 if the currents
go in the same direction through that component or I 1 - I2 if not.)
equation(1):
equation(2):
equation(3):
-𝐸1
+ 𝑅1 𝐼1 + 𝑅2 (𝐼1 + 𝐼2 ) = 0
𝑅4 (𝐼2 + 𝐼3 ) + 𝑅3 𝐼2 + 𝑅2 (𝐼1 + 𝐼2 ) = 0
𝑅4 (𝐼2 + 𝐼3 ) + 𝑅5 𝐼3 − 𝐸2 = 0
4. Solve for unknown loop currents (simultaneous equations). If any solution is negative,
the assumed current direction is wrong.
equation(1):
equation(2):
equation(3):
-20
+ 2𝐼1 + 3𝐼1 + 3𝐼2 = 0
5𝐼2 + 5𝐼3 + 4𝐼2 + 3𝐼1 + 3𝐼2 = 0
5𝐼2 + 5𝐼3 + 6𝐼3 − 5 = 0
equation(1):
equation(2):
equation(3):
5𝐼1 + 3𝐼2
= 20
5𝐼1 + 12𝐼2 + 5𝐼3 = 0
5𝐼2 + 11𝐼3 = 5
Multiply equation (1) by 3 and equation (2) by (-5), then we have:
equation(1):
equation(2):
15𝐼1 + 9𝐼2
= 60
-15𝐼1 − 60𝐼2 − 25𝐼3 = 0
adding these two together gives :
51𝐼2 − 25𝐼3 = 60
Multiply this equation by 11 and equation (3) by 25. You will get:
−561𝐼2 − 275𝐼3 = 660
125𝐼2 + 275𝐼3 = 125
Adding these two will eliminate I3 and give us
−436𝐼2 = 785
𝐼2 = −1.8 A
Back to equation (1) , including in it the calculated value for I2:
5𝐼1 − 3 ∙ 1.8 = 20
which gives us: 𝐼1 = 5.08𝐴
Finally, including this into equation (3) allows us to calculate current I3:
𝐼3 = 1.27A
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DE4401 Lab3 Answers
Using loop currents , we can calculate all branch currents
Ib1 = I1 = 5.08 A
Ib2= I1 + I2 = 3.28 A
Ib3 = -I2 = 1.8 A
Ib4 = I2 + I3 = 0.53 A
Ib5 = I3 = 1.27A
Conclusion (how to the simulation compare with calculation):
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
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DE4401 Lab3 Answers
Branch current method
Steps to follow for the “Branch Current” method of network analysis:
1. Choose a node and assume directions of currents.
2. Write a KCL equation relating currents at the node.
3. Label resistor voltage drop polarities based on assumed currents.
4. Write KVL equations for each loop of the circuit, substituting the product IR for E in
each resistor term of the equations.
5. Solve for unknown branch currents (simultaneous equations).
6. If any solution is negative, then the assumed direction of current for that solution is
wrong.
7. Solve for voltage drops across all resistors (V=IR).
Loop Current Method
Steps to follow for the “Loop (Mesh) Current” method of network analysis:
5. Draw currents in loops of circuit, to account for all components.
6. Label resistor voltage drop polarities based on assumed directions of loop currents.
7. Write KVL equations for each loop, substituting the product IR for E in each resistor
term of the equation. Where two mesh currents intersect through a component, express
the current as the algebraic sum of those two mesh currents (i.e. I1 + I2 if the currents
go in the same direction through that component or I 1 - I2 if not.)
8. Solve for unknown loop currents (simultaneous equations). If any solution is negative,
the assumed current direction is wrong.
9. Algebraically add loop currents to find current in components which share multiple
loop currents.
10. Solve for voltage drops across all resistors (V=IR).
Ohm’s law
Kirchhoff’s current law (KCL)
Kirchhoff’s voltage law (KVL)
Three resistors in series
Three resistors in parallel
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DE4401 Lab3 Answers