Find Io in the circuit in Fig. P2.110.

Name: Problem 110, Chapter 2: Basic Engineering Circuit Analysis
Uploaded: 2021-06-15T19:23:08-08:00
Duration: 5 min 21 s
Channel: Thomas Thompson
Description: Find Io in the circuit in Fig. P2.110.

Key Concepts

Nodal Analysis

Nodal analysis is a method that uses node voltage potentials relative to a common reference point (ground) to analyze circuits. By applying KCL at the nodes, one can formulate equations that allow for the determination of voltage values at different points in the circuit. These nodal voltages can then be used with Ohm’s Law to find branch currents, including the current I?.

Kirchhoff’s Voltage Law (KVL)

KVL is a fundamental principle in circuit analysis that states the algebraic sum of all voltages in any closed loop within a circuit must equal zero. This concept is essential when determining unknown voltages or currents in various parts of a circuit, as it provides a method to relate the voltage drops across passive components like resistors, inductors, and capacitors within the loop.

Kirchhoff’s Current Law (KCL)

KCL is a key rule in analyzing electrical circuits which asserts that the total current entering a node (or junction) must equal the total current leaving the node. This law is crucial in setting up equations that allow for the determination of unknown branch currents, making it indispensable for both nodal analysis and overall circuit analysis.

Ohm’s Law

Ohm’s Law defines the relationship between voltage, current, and resistance in an electrical circuit. It is typically expressed as V = IR, meaning that the voltage drop across a resistor is the product of the current flowing through it and the resistance. This law forms the basis for calculating voltage drops and is fundamental to virtually every circuit analysis technique.

Mesh Analysis

Mesh analysis is a systematic method used to determine the currents flowing in the loops of a planar circuit. By applying KVL around each loop, one can set up and solve a system of linear equations to find the unknown loop currents, which in turn can be used to determine branch currents such as I?. This technique is especially useful in circuits with multiple interconnected loops.

Circuit Simplification Techniques

Circuit simplification techniques, such as Thevenin’s and Norton’s theorems, help reduce complex circuits into simpler equivalent circuits that are easier to analyze. By finding the equivalent voltage source and resistance (or current source and parallel resistance), these methods simplify the process of determining currents and voltages in specific parts of a circuit, which can be particularly useful when calculating I?.

Transcript

00:01 This problem deals with y -dalta transformations.

00:04 It also is going to use oms law, which says the voltage is equal to current times resistance.

00:09 It will also use the formulas that we have for series resistance, which is just the sum of the resistances.

00:20 And for parallel, where it's the reciprocal of the sum of the reciprocals.

00:26 So i've drawn the circuit a little bit differently so that we can see the delta up here.

00:35 And what we're going to do is transform that into a y.

00:39 So here's the way the y delta transformation works.

00:44 I've gone ahead and labeled the resistances here.

00:48 They're all 12 oms.

00:50 But for this top resistance resistance a, then i'm going to multiply r1.

00:58 And r2, and i'm going to divide by the sum of the resistances.

01:03 Then over here for the resistance at b, then i'm going to do the same thing using the resistances on either side of corner b, and the same thing for c.

01:14 So for each of these, since they're all 12 oms, then i have 144 oms divided by the sum of the resistances, that is all three of them.

01:26 And so i have 36 oms and that means that my resistance for each of these for a, b and c, is equal to 4 oms.

01:38 So now i'm going to redraw the circuit again.

01:42 And so my circuit ends up looking like this.

01:45 So i have 4 -oam resistances in each of the branches of that y and i still have the 14 and the 8 down there in the bottom.

01:52 So now i have two branches.

01:54 So here, the resistance of this branch is equal to 4 plus 8, so it's 12 oms.

02:01 Here it's 14 plus 4, so it is 18 oms.

02:07 And now i have a parallel to parallel branches with resistances of 18 and 12...

Please give Ace some feedback