CALCULATIONS Part A. Superposition Principle ð 1 = 1kΩ ð 2 = 2k Solving for ð―ðģ when ðđðģ is infinite (Open Circuit) CALCULATED VALUE OF ð―ðģ WHEN ðđðģ IS INFINITE V1=6V, V2=12V V1=6V, V2=0V V1=0V, V2=12 ððŋ = 8V ððŋ = 4V ððŋ = 4V When ð―ð =6V, ð―ð =12 ð − ð―ðģ ð―ðģ − ðð = +ð ð ð ðð − ðð―ðģ = ð―ðģ − ðð ðð―ðģ = ðð ð―ðģ = ðV When ð―ð =6V, ð―ð =0V ð − ð―ðģ ð―ðģ = +ð ð ð ðð − ðð―ðģ = ð―ðģ ðð―ðģ = ðð ð―ðģ = ðV When ð―ð =0V, ð―ð =12V −ð―ðģ ð―ðģ − ðð = +ð ð ð −ðð―ðģ = ð―ðģ − ðð ðð―ðģ = ðð ð―ðģ = ðV Solving for ð―ðģ when ðđðģ is 3kΩ CALCULATED VALUE OF ð―ðģ WHEN ðđðģ IS 3kΩ V1=6V, V2=12V ððŋ = 6.55V V1=6V, V2=0V ððŋ = 3.27V When ð―ð =6V, ð―ð =12 ð − ð―ðģ ð―ðģ ð―ðģ − ðð = + ð ð ð ðð − ðð―ðģ = ðð―ðģ + ðð―ðģ − ðð ððð―ðģ = ðð ð―ðģ = ð. ððV When ð―ð =6V, ð―ð =0V ð − ð―ðģ ð―ðģ ð―ðģ = + ð ð ð ðð − ðð―ðģ = ðð―ðģ + ðð―ðģ ððð―ðģ = ðð ð―ðģ = ð. ððV When ð―ð =0V, ð―ð =12V −ð―ðģ ð―ðģ ð―ðģ − ðð = + ð ð ð −ðð―ðģ = ðð―ðģ + ðð―ðģ − ðð ððð―ðģ = ðð ð―ðģ = ð. ððV V1=0V, V2=12 ððŋ = 3.27V Solving for Currents ð°ð , ð°ð , ð°ð when ðđðģ is 3kΩ CALCULATED VALUE OF ð°ð , ð°ð , ð°ð WHEN ðđðģ IS 3kΩ V1=6V, V2=12 V1=6V, V2=0V V1=0V, V2=12 ðž1 = 0.55mA ðž1 = −2.73mA ðž1 = 3.27mA ðž2 = −2.73mA ðž2 = 1.64mA ðž2 = −4.37mA ðž3 = 2.18mA ðž3 = 1.09mA ðž3 = 1.09mA When ð―ð =6V, ð―ð =12 ð°ð = ð − ð. ðð ð ð°ð = −ð. ðððĶA ð°ð = ð. ðð − ðð ð ð°ð = −ð. ðððĶA ð°ð = ð. ðð ð ð°ð = ð. ðððĶA When ð―ð =0V, ð―ð =12V ð°ð = ð. ðð ð ð°ð = ð. ðððĶA ð°ð = ð. ðð − ðð ð ð°ð = −ð. ðððĶA ð°ð = ð. ðð ð ð°ð = ð. ðððĶA When ð―ð =6V, ð―ð =0V ð°ð = ð. ðð − ð ð ð°ð = −ð. ðððĶA ð°ð = ð. ðð ð ð°ð = ð. ðððĶA ð. ðð ð ð°ð = ð. ðððĶA ð°ð = Part B. Thevenin Equivalent Circuit Solving for Thevenin Equivalent Circuit when ðđðģ is infinite (Open Circuit) Simulated Values Rth Vth 3kΩ 8V IL 0mA Rth 3kΩ Theoretical Values Vth 8V INSERT EQUIVALENT THEVENIN CIRCUIT HERE When the independent sources are short circuited, ðđðŧð is: ðđðŧð = ð + ð ðđðŧð = ððĪð Solving for ð―ðŧð : ð − ð―ðŧð ð―ðŧð − ðð = +ð ð ð ðð − ðð―ðŧð = ð―ðŧð − ðð ðð―ðŧð = ðð ð―ðŧð = ðV Solving for ð°ðģ : ð°ðģ = ð―ðŧð ðđðŧð + ðđðģ ð°ðģ = ð ð+∞ ð°ðģ = ððĶð IL 0mA Solving for Thevenin Equivalent Circuit when ðđðģ 3kΩ Simulated Values Rth Vth 3kΩ 8V IL 1.3mA Rth 3kΩ Theoretical Values Vth IL 8V 1.33mA INSERT EQUIVALENT THEVENIN CIRCUIT HERE When the independent sources are short circuited, ðđðŧð is: ðđðŧð = ð + ð ðđðŧð = ððĪð Solving for ð―ðŧð : ð − ð―ðŧð ð―ðŧð − ðð = +ð ð ð ðð − ðð―ðŧð = ð―ðŧð − ðð ðð―ðŧð = ðð ð―ðŧð = ðV Solving for ð°ðģ : ð°ðģ = ð―ðŧð ðđðŧð + ðđðģ ð°ðģ = ð ð+ð ð°ðģ = ð. ðððĶð CALCULATED SIMULATED V1=6V, V2=12V V1=6V, V2=0V V1=0V, V2=12V SOLUTION %DIFFERENCE ððŋ = 8V ððŋ = 8V 8−8 × 100 8+8 2 0% ððŋ = 4V ððŋ = 4V 4−4 × 100 4+4 2 0% ððŋ = 4V ððŋ = 4V 4−4 × 100 4+4 2 0% Solving for %Difference of ððŋ (Calculated vs Simulated) when ðđðģ is infinity Solving for %Difference of ððŋ (Calculated vs Simulated) when ðđðģ is 3kΩ CALCULATED SIMULATED V1=6V, V2=12V SOLUTION %DIFFERENCE ððŋ = 6.55V ððŋ = 6.5V | 6.55 − 6.5 | × 100 6.55 + 6.5 2 0.77% ððŋ = 3.27V ððŋ = 3.3V | 3.27 − 3.3 | × 100 3.27 + 3.3 2 0.91% ððŋ = 3.27V ððŋ = 3.3V | 3.27 − 3.3 | × 100 3.27 + 3.3 2 0.91% V1=6V, V2=0V V1=0V, V2=12V Solving for %Difference of ðž1 (Calculated vs Simulated) when ðđðģ is 3kΩ V1=6V, V2=12V CALCULATED SIMULATED ðž1 = 0.55mA ðž1 = 0.6mA SOLUTION %DIFFERENCE 0.55 − 0.6 | × 100 0.55 + 0.6 2 8.70% −2.73 − −2.7 | × 100 −2.73 + −2.7 2 1.10% 3.27 − 3.3 | × 100 3.27 + 3.3 2 0.91% | V1=6V, V2=0V ðž1 = −2.73mA ðž1 = −2.7mA | V1=0V, V2=12V ðž1 = 3.27mA ðž1 = 3.3mA | Solving for %Difference of ðž2 (Calculated vs Simulated) when ðđðģ is 3kΩ CALCULATED SIMULATED V1=6V, V2=12V ðž2 = −2.73mA ðž2 = −2.6mA | V1=6V, V2=0V ðž2 = 1.64mA SOLUTION %DIFFERENCE −2.73 − −2.6 | × 100 −2.73 + −2.6 2 4.88% 1.64 − 1.6 | × 100 1.64 + 1.6 2 2.47% −4.37 − −4.4 | × 100 −4.37 + −4.4 2 0.68% ðž2 = 1.6mA | V1=0V, V2=12V ðž2 = −4.37mA ðž2 = −4.4mA | Solving for %Difference of ðž1 (Calculated vs Simulated) when ðđðģ is 3kΩ CALCULATED SIMULATED SOLUTION %DIFFERENCE | 2.18 − 2.2 | × 100 2.18 + 2.2 2 0.91% | 1.09 − 1.1 | × 100 1.09 + 1.1 2 0.91% V1=6V, V2=12V ðž3 = 2.18mA ðž3 = 2.2mA V1=6V, V2=0V ðž3 = 1.09mA ðž3 = 1.1mA V1=0V, V2=12V ðž3 = 1.09mA ðž3 = 1.1mA | 1.09 − 1.1 | × 100 1.09 + 1.1 2 0.91% Solving for %Difference of Thevenin Equivalent Circuit when ðđðģ is infinite(Open Circuit) CALCULATED SIMULATED SOLUTION %DIFFERENCE | 3−3 | × 100 3+3 2 0% | 8−8 | × 100 8+8 2 0% | 0−0 | × 100 0+0 2 0% ðđðŧð ðđðŧð = ððĪð ðđðŧð = ððĪð ð―ðŧð ð°ðģ ð―ðŧð = ðð ð°ðģ = ððĶð ð―ðŧð = ðð ð°ðģ = ððĶð Solving for %Difference of Thevenin Equivalent Circuit when ðđðģ is infinite(Open Circuit) ðđðŧð ð―ðŧð CALCULATED SIMULATED ðđðŧð = ððĪð ðđðŧð = ððĪð ð―ðŧð = ðð SOLUTION %DIFFERENCE | 3−3 | × 100 3+3 2 0% | 8−8 | × 100 8+8 2 0% ð―ðŧð = ðð ð°ðģ ð°ðģ = ð. ðððĶð ð°ðģ = ð. ððĶð | 1.33 − 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.
