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CALCULATIONS

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CALCULATIONS
Part A. Superposition Principle
𝑅1 = 1kΞ©
𝑅2 = 2k
Solving for 𝑽𝑳 when 𝑹𝑳 is infinite (Open Circuit)
CALCULATED VALUE OF 𝑽𝑳 WHEN 𝑹𝑳 IS INFINITE
V1=6V, V2=12V
V1=6V, V2=0V
V1=0V, V2=12
𝑉𝐿 = 8V
𝑉𝐿 = 4V
𝑉𝐿 = 4V
When π‘½πŸ =6V, π‘½πŸ =12
πŸ” βˆ’ 𝑽𝑳 𝑽𝑳 βˆ’ 𝟏𝟐
=
+𝟎
𝟏
𝟐
𝟏𝟐 βˆ’ πŸπ‘½π‘³ = 𝑽𝑳 βˆ’ 𝟏𝟐
πŸ‘π‘½π‘³ = πŸπŸ’
𝑽𝑳 = πŸ–V
When π‘½πŸ =6V, π‘½πŸ =0V
πŸ” βˆ’ 𝑽𝑳 𝑽𝑳
=
+𝟎
𝟏
𝟐
𝟏𝟐 βˆ’ πŸπ‘½π‘³ = 𝑽𝑳
πŸ‘π‘½π‘³ = 𝟏𝟐
𝑽𝑳 = πŸ’V
When π‘½πŸ =0V, π‘½πŸ =12V
βˆ’π‘½π‘³ 𝑽𝑳 βˆ’ 𝟏𝟐
=
+𝟎
𝟏
𝟐
βˆ’πŸπ‘½π‘³ = 𝑽𝑳 βˆ’ 𝟏𝟐
πŸ‘π‘½π‘³ = 𝟏𝟐
𝑽𝑳 = πŸ’V
Solving for 𝑽𝑳 when 𝑹𝑳 is 3kΞ©
CALCULATED VALUE OF 𝑽𝑳 WHEN 𝑹𝑳 IS 3kΞ©
V1=6V, V2=12V
𝑉𝐿 = 6.55V
V1=6V, V2=0V
𝑉𝐿 = 3.27V
When π‘½πŸ =6V, π‘½πŸ =12
πŸ” βˆ’ 𝑽𝑳 𝑽𝑳 𝑽𝑳 βˆ’ 𝟏𝟐
=
+
𝟏
πŸ‘
𝟐
πŸ‘πŸ” βˆ’ πŸ”π‘½π‘³ = πŸπ‘½π‘³ + πŸ‘π‘½π‘³ βˆ’ πŸ‘πŸ”
πŸπŸπ‘½π‘³ = πŸ•πŸ
𝑽𝑳 = πŸ”. πŸ“πŸ“V
When π‘½πŸ =6V, π‘½πŸ =0V
πŸ” βˆ’ 𝑽𝑳 𝑽𝑳 𝑽𝑳
=
+
𝟏
πŸ‘
𝟐
πŸ‘πŸ” βˆ’ πŸ”π‘½π‘³ = πŸπ‘½π‘³ + πŸ‘π‘½π‘³
πŸπŸπ‘½π‘³ = πŸ‘πŸ”
𝑽𝑳 = πŸ‘. πŸπŸ•V
When π‘½πŸ =0V, π‘½πŸ =12V
βˆ’π‘½π‘³ 𝑽𝑳 𝑽𝑳 βˆ’ 𝟏𝟐
=
+
𝟏
πŸ‘
𝟐
βˆ’πŸ”π‘½π‘³ = πŸπ‘½π‘³ + πŸ‘π‘½π‘³ βˆ’ πŸ‘πŸ”
πŸπŸπ‘½π‘³ = πŸ‘πŸ”
𝑽𝑳 = πŸ‘. πŸπŸ•V
V1=0V, V2=12
𝑉𝐿 = 3.27V
Solving for Currents π‘°πŸ , π‘°πŸ , π‘°πŸ‘ when 𝑹𝑳 is 3kΞ©
CALCULATED VALUE OF π‘°πŸ , π‘°πŸ , π‘°πŸ‘ WHEN 𝑹𝑳 IS 3kΞ©
V1=6V, V2=12
V1=6V, V2=0V
V1=0V, V2=12
𝐼1 = 0.55mA
𝐼1 = βˆ’2.73mA
𝐼1 = 3.27mA
𝐼2 = βˆ’2.73mA
𝐼2 = 1.64mA
𝐼2 = βˆ’4.37mA
𝐼3 = 2.18mA
𝐼3 = 1.09mA
𝐼3 = 1.09mA
When π‘½πŸ =6V, π‘½πŸ =12
π‘°πŸ =
πŸ” βˆ’ πŸ”. πŸ“πŸ“
𝟏
π‘°πŸ = βˆ’πŸŽ. πŸ“πŸ“π¦A
π‘°πŸ =
πŸ”. πŸ“πŸ“ βˆ’ 𝟏𝟐
𝟐
π‘°πŸ = βˆ’πŸ. πŸ•πŸ‘π¦A
π‘°πŸ‘ =
πŸ”. πŸ“πŸ“
πŸ‘
π‘°πŸ‘ = 𝟐. πŸπŸ–π¦A
When π‘½πŸ =0V, π‘½πŸ =12V
π‘°πŸ =
πŸ‘. πŸπŸ•
𝟏
π‘°πŸ = πŸ‘. πŸπŸ•π¦A
π‘°πŸ =
πŸ‘. πŸπŸ• βˆ’ 𝟏𝟐
𝟐
π‘°πŸ = βˆ’πŸ’. πŸ‘πŸ•π¦A
π‘°πŸ‘ =
πŸ‘. πŸπŸ•
πŸ‘
π‘°πŸ‘ = 𝟏. πŸŽπŸ—π¦A
When π‘½πŸ =6V, π‘½πŸ =0V
π‘°πŸ =
πŸ‘. πŸπŸ• βˆ’ πŸ”
𝟏
π‘°πŸ = βˆ’πŸ. πŸ•πŸ‘π¦A
π‘°πŸ =
πŸ‘. πŸπŸ•
𝟐
π‘°πŸ = 𝟏. πŸ”πŸ’π¦A
πŸ‘. πŸπŸ•
πŸ‘
π‘°πŸ‘ = 𝟏. πŸŽπŸ—π¦A
π‘°πŸ‘ =
Part B. Thevenin Equivalent Circuit
Solving for Thevenin Equivalent Circuit when 𝑹𝑳 is infinite (Open Circuit)
Simulated Values
Rth
Vth
3kΞ©
8V
IL
0mA
Rth
3kΞ©
Theoretical Values
Vth
8V
INSERT EQUIVALENT THEVENIN CIRCUIT HERE
When the independent sources are short circuited, 𝑹𝑻𝒉 is:
𝑹𝑻𝒉 = 𝟏 + 𝟐
𝑹𝑻𝒉 = πŸ‘π€π›€
Solving for 𝑽𝑻𝒉 :
πŸ” βˆ’ 𝑽𝑻𝒉 𝑽𝑻𝒉 βˆ’ 𝟏𝟐
=
+𝟎
𝟏
𝟐
𝟏𝟐 βˆ’ πŸπ‘½π‘»π’‰ = 𝑽𝑻𝒉 βˆ’ 𝟏𝟐
πŸ‘π‘½π‘»π’‰ = πŸπŸ’
𝑽𝑻𝒉 = πŸ–V
Solving for 𝑰𝑳 :
𝑰𝑳 =
𝑽𝑻𝒉
𝑹𝑻𝒉 + 𝑹𝑳
𝑰𝑳 =
πŸ–
πŸ‘+∞
𝑰𝑳 = πŸŽπ¦π€
IL
0mA
Solving for Thevenin Equivalent Circuit when 𝑹𝑳 3kΞ©
Simulated Values
Rth
Vth
3kΞ©
8V
IL
1.3mA
Rth
3kΞ©
Theoretical Values
Vth
IL
8V
1.33mA
INSERT EQUIVALENT THEVENIN CIRCUIT HERE
When the independent sources are short circuited, 𝑹𝑻𝒉 is:
𝑹𝑻𝒉 = 𝟏 + 𝟐
𝑹𝑻𝒉 = πŸ‘π€π›€
Solving for 𝑽𝑻𝒉 :
πŸ” βˆ’ 𝑽𝑻𝒉 𝑽𝑻𝒉 βˆ’ 𝟏𝟐
=
+𝟎
𝟏
𝟐
𝟏𝟐 βˆ’ πŸπ‘½π‘»π’‰ = 𝑽𝑻𝒉 βˆ’ 𝟏𝟐
πŸ‘π‘½π‘»π’‰ = πŸπŸ’
𝑽𝑻𝒉 = πŸ–V
Solving for 𝑰𝑳 :
𝑰𝑳 =
𝑽𝑻𝒉
𝑹𝑻𝒉 + 𝑹𝑳
𝑰𝑳 =
πŸ–
πŸ‘+πŸ‘
𝑰𝑳 = 𝟏. πŸ‘πŸ‘π¦π€
CALCULATED SIMULATED
V1=6V, V2=12V
V1=6V, V2=0V
V1=0V, V2=12V
SOLUTION
%DIFFERENCE
𝑉𝐿 = 8V
𝑉𝐿 = 8V
8βˆ’8
× 100
8+8
2
0%
𝑉𝐿 = 4V
𝑉𝐿 = 4V
4βˆ’4
× 100
4+4
2
0%
𝑉𝐿 = 4V
𝑉𝐿 = 4V
4βˆ’4
× 100
4+4
2
0%
Solving for %Difference of 𝑉𝐿 (Calculated vs Simulated) when 𝑹𝑳 is infinity
Solving for %Difference of 𝑉𝐿 (Calculated vs Simulated) when 𝑹𝑳 is 3kΞ©
CALCULATED SIMULATED
V1=6V, V2=12V
SOLUTION
%DIFFERENCE
𝑉𝐿 = 6.55V
𝑉𝐿 = 6.5V
|
6.55 βˆ’ 6.5
| × 100
6.55 + 6.5
2
0.77%
𝑉𝐿 = 3.27V
𝑉𝐿 = 3.3V
|
3.27 βˆ’ 3.3
| × 100
3.27 + 3.3
2
0.91%
𝑉𝐿 = 3.27V
𝑉𝐿 = 3.3V
|
3.27 βˆ’ 3.3
| × 100
3.27 + 3.3
2
0.91%
V1=6V, V2=0V
V1=0V, V2=12V
Solving for %Difference of 𝐼1 (Calculated vs Simulated) when 𝑹𝑳 is 3kΞ©
V1=6V, V2=12V
CALCULATED
SIMULATED
𝐼1 = 0.55mA
𝐼1 = 0.6mA
SOLUTION
%DIFFERENCE
0.55 βˆ’ 0.6
| × 100
0.55 + 0.6
2
8.70%
βˆ’2.73 βˆ’ βˆ’2.7
| × 100
βˆ’2.73 + βˆ’2.7
2
1.10%
3.27 βˆ’ 3.3
| × 100
3.27 + 3.3
2
0.91%
|
V1=6V, V2=0V
𝐼1 = βˆ’2.73mA 𝐼1 = βˆ’2.7mA
|
V1=0V, V2=12V
𝐼1 = 3.27mA
𝐼1 = 3.3mA
|
Solving for %Difference of 𝐼2 (Calculated vs Simulated) when 𝑹𝑳 is 3kΞ©
CALCULATED
SIMULATED
V1=6V, V2=12V 𝐼2 = βˆ’2.73mA 𝐼2 = βˆ’2.6mA
|
V1=6V, V2=0V
𝐼2 = 1.64mA
SOLUTION
%DIFFERENCE
βˆ’2.73 βˆ’ βˆ’2.6
| × 100
βˆ’2.73 + βˆ’2.6
2
4.88%
1.64 βˆ’ 1.6
| × 100
1.64 + 1.6
2
2.47%
βˆ’4.37 βˆ’ βˆ’4.4
| × 100
βˆ’4.37 + βˆ’4.4
2
0.68%
𝐼2 = 1.6mA
|
V1=0V, V2=12V 𝐼2 = βˆ’4.37mA 𝐼2 = βˆ’4.4mA
|
Solving for %Difference of 𝐼1 (Calculated vs Simulated) when 𝑹𝑳 is 3kΞ©
CALCULATED
SIMULATED
SOLUTION
%DIFFERENCE
|
2.18 βˆ’ 2.2
| × 100
2.18 + 2.2
2
0.91%
|
1.09 βˆ’ 1.1
| × 100
1.09 + 1.1
2
0.91%
V1=6V, V2=12V 𝐼3 = 2.18mA 𝐼3 = 2.2mA
V1=6V, V2=0V
𝐼3 = 1.09mA 𝐼3 = 1.1mA
V1=0V, V2=12V 𝐼3 = 1.09mA 𝐼3 = 1.1mA
|
1.09 βˆ’ 1.1
| × 100
1.09 + 1.1
2
0.91%
Solving for %Difference of Thevenin Equivalent Circuit when 𝑹𝑳 is infinite(Open Circuit)
CALCULATED
SIMULATED
SOLUTION
%DIFFERENCE
|
3βˆ’3
| × 100
3+3
2
0%
|
8βˆ’8
| × 100
8+8
2
0%
|
0βˆ’0
| × 100
0+0
2
0%
𝑹𝑻𝒉 𝑹𝑻𝒉 = πŸ‘π€π›€ 𝑹𝑻𝒉 = πŸ‘π€π›€
𝑽𝑻𝒉
𝑰𝑳
𝑽𝑻𝒉 = πŸ–π•
𝑰𝑳 = πŸŽπ¦π€
𝑽𝑻𝒉 = πŸ–π•
𝑰𝑳 = πŸŽπ¦π€
Solving for %Difference of Thevenin Equivalent Circuit when 𝑹𝑳 is infinite(Open Circuit)
𝑹𝑻𝒉
𝑽𝑻𝒉
CALCULATED
SIMULATED
𝑹𝑻𝒉 = πŸ‘π€π›€
𝑹𝑻𝒉 = πŸ‘π€π›€
𝑽𝑻𝒉 = πŸ–π•
SOLUTION
%DIFFERENCE
|
3βˆ’3
| × 100
3+3
2
0%
|
8βˆ’8
| × 100
8+8
2
0%
𝑽𝑻𝒉 = πŸ–π•
𝑰𝑳
𝑰𝑳 = 𝟏. πŸ‘πŸ‘π¦π€ 𝑰𝑳 = 𝟏. πŸ‘π¦π€
|
1.33 βˆ’ 1.3
| × 100
1.33 + 1.3
2
2.28%
How well the measurements agree with the superposition principle? (give differences in %).
Does the Thevenin equivalent circuit provide the same voltage across a load as the circuit it
substitutes? If not explain possible reasons for discrepancies. Do not forget to number figures
and tables and to give them captions (titles). Number all pages of the report.
-
In our activity, we simulated and calculated the VL and the currents across the resistor, given
the value of the RL is 3000 ohms and infinite by using superposition and Thevenin’s method.
The table 1.1 is where the value of the RL is infinite, while table 1.2 is where we set the RL
to 3000 ohms. Then table 1.3 is where the values of the of the current across each resistor
where the RL is 3000 ohms. Then the group calculated the percent difference between
simulated and calculated values and presented it in table 4.1- 4.7. The range of the percent
difference are all below 10% which is a good indication that the values are close to each
other.
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