Electrical Circuit I Assignment 2 (2024) - PDF

# Electrical Circuit I ## Assignment 2 - Date: 18/3/2024 ## Course Details - **Course Code:** ELP 113 - **Instructor:** Assist. Prof. Magda Ibrahim - **Level/Year:** 100-2023/2024 - **Student Name:** Ali Maher darwish - **I.D. #:** 270 206 ## Question #1 ### [SO 1] For the circuit shown, determi...

# Electrical Circuit I ## Assignment 2 - Date: 18/3/2024 ## Course Details - **Course Code:** ELP 113 - **Instructor:** Assist. Prof. Magda Ibrahim - **Level/Year:** 100-2023/2024 - **Student Name:** Ali Maher darwish - **I.D. #:** 270 206 ## Question #1 ### [SO 1] For the circuit shown, determine a numerical value for the voltage labeled $v_1$ using nodal analysis ## Circuit Diagram The circuit is comprised of the following: - Resistors: 1 $\Omega$, 5$\Omega$, 9$\Omega$ - Voltage source: 9V - Current sources: 3A, 5A The provided diagram is a rectangular circuit with the following components: - Top: 1 $\Omega$ resistor, labeled with voltage $v_2$. - Left: 3A current source flowing upwards. - Right: 5A current source flowing downwards. - Bottom: 9V voltage source connected to the node between the 5$\Omega$ and 9$\Omega$ resistors. - Bottom Left: 5$\Omega$ resistor connected between the node on the left and the top of the voltage source. - Bottom Right: 9$\Omega$ resistor connected between the node on the right and the top of the voltage source. - Voltage $v_1$ is measured across the node on the left and the node on the right. ## Nodal Analysis Equation - The node between the 5$\Omega$ and 9$\Omega$ resistors is designated as a supernode. - The total current entering the supernode is equal to the total current leaving it. - The current entering the supernode can be written as $I_1 + I_2 + 5$. - The current leaving the supernode can be written as $I_1 + I_2$. - Therefore the first nodal equation is: $$3 = I_1 + I_2 + 5$$. - The current through the 5$\Omega$ resistor is $I_1 = \frac{V_1}{5}$. - The current through the 9$\Omega$ resistor is $I_2 = \frac{V_2}{9}$. - The voltage across the voltage source can be written as $V_1 - V_2 = 9$. - Combining these, we obtain a second nodal equation: $$ -2 = \frac{V_1}{5} + \frac{V_2}{9}$$. - We can multiply this equation by 45 to get: $$ -90 = 9V_1 + 5V_2$$. - Using the equations provided, we can solve for the voltages $V_1$ and $V_2$. - The solutions are: $$V_1 = -3.21 V$$ $$V_2 = -12.21 V$$. ## Conclusion - The numerical value for the voltage labeled $v_1$ is **-3.21V**. ## Acknowledgement With all best wishes Assist. Prof. Magda Ebrahim