{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

nodal_analysis_problem1 - Since the ideal source is a...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Sample Problem Topic: Voltage & Current Dividers Statement of Problem: Given the circuit shown in the figure below Find a) The value of currents i 1 and i 2 b) The voltage of v 0 Solution We will use the Node Voltage/Nodal Analysis approach to solve this problem. Define v 1 : 90 Ω 4A 80 Ω i 2 i 1 30 Ω v 1 Create the linear equation. The equation is based using the bottom node as ground, or “0.” 4 + 𝑣 1 80 + 𝑣 1 30 + 90 = 0 Each component is defined by following the v 1 node to ground. Looking closely, we are taking a voltage, and dividing by a resistance. Using Ohm’s law, if we inspect this more closely, we are defining currents. V = IR, rearranging this equation, we get: V/R = I
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Since the ideal source is a current sou rce, and it is going “against” the direction from v 1 to ground, we defined it as a negative value. Solving the linear equation, we find v 1 . Note: V 1 = V c) V 1 = 192V d) Knowing the voltage at the node, and knowing that parallel resistors share the same voltage drop, we can simply use Ohm’s law to determine the currents through each branch. V = I *R 192V = I * 80 Ω 192V = I * (30 Ω +90 Ω ) b) I 2 = 1.6A a) I 1 = 2.4A...
View Full Document

{[ snackBarMessage ]}