Electricity By using Kirchhoff's Laws, it can be shown that...

Electricity By using Kirchhoff's Laws, it can be shown that the currents $I_{1} I_{2},$ and $I_{3}$ that pass through the three branches of the circuit in the figure satisfy the given linear system. Solve the system to find $I_{1}, I_{2},$ and $I_{3}$ . $$ \left\{\begin{aligned} I_{1}+I_{2}-I_{3} &=0 \\ 16 I_{1}-8 I_{2} &=4 \\ 8 I_{2}+4 I_{3} &=5 \end{aligned}\right. $$

Electricity By using Kirchhoff's Laws, it can be shown that
the currents $I_{1} I_{2},$ and $I_{3}$ that pass through the three branches
of the circuit in the figure satisfy the given linear system.
Solve the system to find $I_{1}, I_{2},$ and $I_{3}$ .
$$
\left\{\begin{aligned} I_{1}+I_{2}-I_{3} &=0 \\ 16 I_{1}-8 I_{2} &=4 \\ 8 I_{2}+4 I_{3} &=5 \end{aligned}\right.
$$

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Key Concepts

Kirchhoff's Circuit Laws

Kirchhoff's Circuit Laws, including the Current Law (KCL) and Voltage Law (KVL), are fundamental principles in electrical circuits. KCL states that the total current entering a junction equals the total current leaving it, while KVL states that the sum of the potential differences (voltages) around any closed loop in a circuit must equal zero. These laws provide the basis for writing the equations that describe the behavior of electrical circuits.

Systems of Linear Equations

Systems of linear equations consist of multiple linear equations with common variables. In the context of circuit analysis, Kirchhoff's Laws lead to a system of equations that represent the relationships between the various currents and voltages in the circuit. Solving these systems is essential to determine the electrical quantities in different branches of the circuit.

Algebraic Solution Methods

Algebraic solution methods, such as substitution and elimination, are used to solve systems of linear equations. These techniques involve manipulating the equations to isolate and solve for the unknown variables. In circuit analysis, these methods are applied to the equations derived from Kirchhoff's Laws to compute the current or voltage values in the circuit.

Transcript

00:01 So by using church offs law, it can be shown that the current i -1, i -2, and i -3 that pass through the three branches of the circuit and the figure satisfied a given system in your system.

00:14 And now we're asked to solve for i -1, i -2, and i -3.

00:19 Okay.

00:20 This is i -1, i -2, and i -3.

00:24 Okay, so now looking at the three equations that we have, equation one, two, and three.

00:38 Let's see what i can do.

00:41 I2 in terms of i -1.

00:50 How she would start off.

00:52 So our first equation has three terms, i -1, i -2, i -3.

00:58 Our second equation has i -1 and i -2, and our third equation has i -2 and i -3.

01:07 Okay.

01:09 Well, if you look over here, we can rewrite our second equation in terms of i1 and then plug i1 into equation 1 and then we'll have an equation in terms of i 2 and i 1 i 3 and then we can use equation 1 and equation 3 to solve for either i 2 or i 3 so let's do that i have 16 i 1 and then i'm going to add 8 i 2 to both sides so i get 4 plus 8 i2.

01:49 Now i'm going to divide by 16 on both sides.

01:52 I get i1 is equal to 4 over 16, which is 1 over 4, plus 8 over 16.

02:05 That's just 1 half, i2.

02:08 Ok, now i'm going to plug i1 into equation 1...

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