If VR=15 V, find Vx in Fig. P 2.91.

Name: Problem 91, Chapter 2: Basic Engineering Circuit Analysis
Uploaded: 2021-06-10T17:29:06-08:00
Duration: 5 min 41 s
Channel: Thomas Thompson
Description: If VR=15 V, find Vx in Fig. P 2.91.

Key Concepts

Kirchhoff's Voltage Law

Kirchhoff's Voltage Law (KVL) states that the algebraic sum of all voltages in any closed loop of a circuit must equal zero. This principle is used to relate the voltages across different elements in the loop, allowing for the calculation of unknown voltages when other voltages are known.

Ohm's Law

Ohm's Law defines the relationship between voltage, current, and resistance in a circuit through the formula V = IR. This law is critical for determining the voltage drop across resistive elements and forms the basis of many circuit analysis techniques.

Voltage Divider Rule

The Voltage Divider Rule is a method used in series circuits to determine the voltage across a particular resistor. It involves calculating the proportion of the total voltage that appears across the resistor based on its resistance relative to the total resistance in the series circuit.

Transcript

00:01 This problem uses kirchhoff's current law, which says that the sum of the currents at any particular note is equal to zero, or that the sum of the currents going in is equal to the sum of the currents going out.

00:16 Also going to use holmes law.

00:20 It says that voltage is equal to current times resistance, and also the understanding that for a parallel circuit, that the voltages across the branches are equal.

00:31 So i'm going to work from this voltage drop across the 5 -oom resistor being 15 volts all the way back here to find out what v .sub -x is.

00:42 And i'll start by applying kirchhoff's current law at node a, and i'm going to do that for each of these nodes.

00:49 At node a, going out, i have i -1 plus i -2, and then going in there from that current supply i have three amps.

01:00 Then at b, i have i2 coming in, and that's equal to i3 plus i4.

01:12 Then at c, i have i4 going in and then going out, i have two amps, and i have i sub five.

01:23 Now, i also know that over here, using this, that voltage is equal to, voltage at that resistor is equal to i5 times that resistance of 5 oms.

01:40 And that's equal to 15 volts.

01:42 That means that i5 is equal to 3, not oms.

01:48 That would be ridiculous.

01:50 It's equal to 3 amps.

01:51 It being three amps, that also means that three amps is going through that three -oam resistor.

01:58 So the voltage from c to e, that is from here to here, that voltage drop across that branch, and therefore every other branch is going to be equal to i5, which is three amps, times the total resistance, and i just have three oms and five oms in series.

02:19 And so it's going to be equal to 24 volts.

02:24 So that's my voltage drop across each of those.

02:28 And so what i'm going to do now is work.

02:31 And so if that is the case, then i know that i4 is equal to 5 amps.

02:41 Because there at node c, i had 2 amps going out.

02:45 I have 3 amps going out through that last branch.

02:49 And so i4 is going to be equal to 5 amps...

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