Using Kirchhoff's Laws (TAGALOG) (English Sub)

Introduction to Electric Networks00:00:02

An electric network is composed of various elements connected together. Key components include nodes (points where two or more elements connect, preferably three or more), branches (parts of the network between two junctions), and loops (closed paths starting and ending at the same point).

Kirchhoff's Laws: KCL and KVL00:03:56

The video introduces Kirchhoff's Current Law (KCL), stating that the total current entering a node equals the total current leaving it. Kirchhoff's Voltage Law (KVL) is then explained, which states that the algebraic sum of products of current and resistance (IR) and electromotive forces (EMFs) in any closed loop is zero. Guidance is provided on determining the sign convention for voltage sources (+E or -E) and IR drops (+IR or -IR) based on the loop direction relative to the component or current.

Assuming Current Directions00:06:57

When current directions are not given, they can be assumed. If a calculated current value is positive, the assumed direction is correct. If it's negative, the actual current direction is opposite to the assumed one.

Example: Setting up Equations using KCL and KVL00:07:53

A circuit analysis example begins by assuming directions for three unknown currents (I1, I2, I3) and establishing two clockwise loops. Three equations are needed to solve for these unknowns: one from KCL at a node and two from KVL for each loop.

Applying KCL at Node A00:09:01

At node A, where I2 enters and I1 and I3 leave, the KCL equation is derived as I2 = I1 + I3. This becomes the first equation for solving the circuit.

Applying KVL to Loop 100:09:26

For Loop 1 (clockwise), the KVL equation is formulated by considering voltage drops and rises. The equation derived is 2I1 + I2 - 4 = 0, which serves as the second equation.

Applying KVL to Loop 200:11:15

For Loop 2 (clockwise), the KVL equation is similarly formulated. The terms involve voltage sources and IR drops, resulting in the third equation: -I2 - 10I3 - 8 = 0.

Solving the System of Equations00:12:27

With three equations established, algebraic methods are used to solve for I1, I2, and I3. The video demonstrates substituting Equation 1 (I2 = I1 + I3) into Equations 2 and 3 to reduce the number of unknowns. Then, the elimination method is applied to solve for the remaining variables.

Determining Current Values and Directions00:15:24

After solving, the values found are I1 = 1.625 A, I2 = 0.75 A, and I3 = -0.875 A. The negative sign for I3 indicates that its actual direction is opposite to the assumed direction, while the positive values for I1 and I2 mean their assumed directions are correct.