Electricity: Kirchhofr's Rules An application of Kirchhoff's...

Electricity: Kirchhofr's Rules An application of Kirchhoff's Rules to the circuit shown results in the following system of equations:
$$
\left\{\begin{aligned}
I_1 & =I_3+I_2 \\
24-6 I_1-3 I_3 & =0 \\
12+24-6 I_1-6 I_2 & =0
\end{aligned}\right.
$$
Find the currents $I_1 I_2$, and $I_2$.

Key Concepts

Kirchhoff's Current Law (KCL)

KCL is a fundamental principle stating that the sum of currents entering a node (or junction) in a circuit is equal to the sum of currents leaving that node. This law is used to establish relationships between different branch currents in the circuit, ensuring that charge is conserved at every junction.

Kirchhoff's Voltage Law (KVL)

KVL asserts that within any closed loop in an electrical circuit, the algebraic sum of all the voltage rises and drops must equal zero. This principle is key in formulating equations based on the potential differences across various components in the circuit.

Mesh (Loop) Analysis

Mesh analysis is a systematic method of applying KVL to individual loops in a circuit to develop a set of simultaneous equations. By assigning loop currents and applying KVL along each loop, one can analyze the circuit’s behavior and solve for unknown currents.

Solving Systems of Linear Equations

In circuit analysis, the equations derived from KCL and KVL form a system of linear equations. Techniques such as substitution, elimination, or matrix methods are used to find the numerical values of the unknown variables, such as currents and voltages, ensuring the circuit obeys both Kirchhoff's rules.

Transcript

00:02 So in this problem, we're given a circuit, and we're given that by kirchhoff's rules, the current around the different circuits gives us these equations.

00:15 I3 equals i1 plus i2, where each of these is the circuit, whereas the amps, the flow, the current, excuse me, current, in each of those circuits.

00:27 And 8 is 4 i3 plus 6 i2 and 8 i1 is 4 plus 6 i2 okay so we're given these three equations with three unknowns call this equation 1 this equation 2 this equation 3 and i want to write these a little bit neater here.

01:02 Okay.

01:03 So all i'm going to do is rewrite these so that i have the constant equals something i want, something i two, something i three.

01:13 Okay.

01:14 So on the first equation, i mean they have zero equals i1 plus i2 minus i3.

01:24 That's equation one.

01:26 I have eight equals there is no i1 term and it's 6 i2 plus 4 i3 equation 2 and equation 3 well i move the 4 to the other side so that's minus 4 and the 8 i1 to the other side so that's minus 8 i 1 plus 6 i 2 and there is no i 3 in this equation alright, so now what can i see? well, i can see that if i put equations 1 and 3 together, then i could get the i1s to drop out, and i2 is already got nothing for i1.

02:30 So that would give me two equations and two unknowns.

02:34 Okay, so we'll go 8 times, equation 1, so that's 0, equals 8 i1 plus 8 i2, plus 8, oops, not plus, minus 8 i3.

03:02 Okay, and i'm going to add equation 3 to that.

03:06 That's minus 4 is minus 8 i1, plus 6i2, plus 0, and and the i1s will drop out.

03:23 So i'm left with minus 4 is, well, 8 plus 6, that's 14i2, minus 8, i3.

03:38 And i'm going to call this equation 4.

03:42 Okay, so now what can i see between equations 2 and 4? well, if i multiply equation 2 by 2 by the number 2, then the i3s will drop out.

03:56 So 2 times equation 2 gives me 16 is 12 i2 plus 8 i3.

04:12 And of course i'm going to add equation 4 of that minus 4 equals 14 i2 minus 8 i3 and the i 3 drop out.

04:31 So what am i left with? i'm left with 12 equals 26 i2.

04:39 Okay.

04:41 So divide by 26, both of those 12 and 26 are divisible by 2...

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