{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}


current_voltage_divider_problem3 - Redraw the circuit as...

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: The value of currents i 0 , i 1 and i 2 Solution To begin to solve this circuit, we need to know the value of v . This voltage will determine the dependent current source value. We can then use current and voltage dividers to solve the remainder of the circuit. i 0 Note: We know via KCL/KVL that all current coming into a node must leave the node. Since I 0 is coming into a node connecting two different parts of the circuit, we also can assume that it is leaving the node. This implies that there is 0 current flow, thus causing i 0 to be 0. Using a voltage divider to solve for v : v = 60 𝑉 ∗ 5000 5000 + 1000 = 50 𝑉 This determines that the dependent current source is: 6 * 10 -3 * (50) = .3A Knowing that i 0 = 0A, we can consider the right half of the circuit only.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Redraw the circuit as follows: 500Ω .3A 2kΩ i 2 i 1 We can calculate the current through the parallel circuit, thus obtaining the amount of current split through the 2k Ω and 500Ω resistors Simply use the current divider equation to get the final current division. ² 1 = .3 ³ ∗ 500 2000 + 500 ± = − .06 ³ ´µ − 60 ¶³ ² 2 = .3 ³ ∗ 2000 2000 + 500 ± = − .24 ³ ´µ − 240 ¶³ Since the current source is pointing downwards, this defines the convention of the current flow. When looking at the top node, you will notice that the arrows of i1 and i2 are pointing “outwards.” Notice that the positive current direction is outwards from the top node as well. Since all currents going in or out of a node sum to zero, it explains why the currents are negative....
View Full Document

{[ snackBarMessage ]}