Consider the circuit shown in Figure P 28.9 . Find (a) the current in the 20.0-Ωresistor and (b)

Consider the circuit shown in Figure P 28.9 . Find (a) the...

Key Concepts

Series and Parallel Resistor Networks

Understanding series and parallel resistor networks is key to circuit analysis. In a series configuration, resistors add directly, affecting the overall resistance and current flow. In parallel configurations, the inverse of the total resistance is the sum of the inverses of the individual resistances. Recognizing these arrangements allows for the simplification of complex circuits into more manageable equivalents, facilitating easier computation of voltage and current.

Voltage Divider Principle

The voltage divider principle is a useful concept for determining the voltage across a particular resistor in a series circuit. It states that the voltage across a resistor is proportional to its resistance relative to the total resistance in the series. This concept is particularly useful when needing to find the potential difference between two points in a series network, as it directly relates the resistor values to the voltage distribution.

Ohm's Law

Ohm's Law is a fundamental principle in electrical circuits that relates voltage (V), current (I), and resistance (R) through the equation V = IR. This law is essential for predicting the behavior of circuits, allowing one to calculate the potential difference across an element if the current and resistance are known, or vice versa.

Kirchhoff's Circuit Laws

Kirchhoff's Circuit Laws, which include the Voltage Law (KVL) and the Current Law (KCL), are critical for analyzing complex circuits. KVL states that the sum of potential differences around any closed loop in a circuit must equal zero, while KCL asserts that the total current entering a junction must equal the total current leaving it. These laws help in setting up equations that govern the flow of current and the distribution of voltage in a circuit.

Transcript

00:01 Okay, so in this problem, we are given a circuit that has five resistors, one voltage source, and we're asked to find the current and the potential difference at different points in the circuit.

00:13 Specifically, they want to know what the current is through this, what i've called r1, this resistor here, and what the voltage difference is between points a and b.

00:24 So the way i always teach to do this is to just analyze the circuit.

00:28 That way, it doesn't matter what specific resistor they're asking about you can find the current through it.

00:36 And it's all in one table.

00:38 So what i've done here is i wrote out a table of all the different voltages, currents, and resistances for each of the five components as well as the total.

00:49 So the first step in trying to solve these problems is always going to be, or is usually going to be finding this total resistance here.

00:56 So i need to find the total resistance of this circuit.

01:00 If we look here, we notice that there is a resistor that is in series with everything else, and that is resistor 5 here, meaning that no matter which loop i go to reach the other end of the battery, i always have to go through resistor 5.

01:19 And so that means that r5 is in series with the battery.

01:24 However, this section here, this r1 and 2, and r3 and r4, are in parallel with each other.

01:34 So if we just draw a really simple sort of redraw of this, because i always think it'll make it a little bit easier to visualize, we have, let's actually redraw like this, we have r1 here, which goes into r2, which runs into r5.

01:55 So this is r1, r2, and r5.

01:59 Then we have another branch that goes like this.

02:02 It's r3, and one more that goes like this.

02:05 That's r4.

02:07 Now i re -drew it this way because to me and to most of my students, this is like an easier way to visualize it.

02:12 First, i'm going to combine these two because they're in series, and then i'm going to combine all three of those because they're in parallel.

02:20 So let's combine r1 and two first.

02:21 So now we have a circuit that looks something like this, where this equivalent resistance right here is the combination of 1 and 2, which is 25.

02:38 So this is a 25 oms.

02:40 R3 is a 5 oms, and r4 is a 10 oms.

02:44 And then we have this final 10 ome in series with everything.

02:48 So let's combine these three into an equivalent resistance.

02:51 To do that, we're going to use our equivalent resistance equation.

02:54 So 1 over the equivalent resistance is equal to 1 over the first resistor, so 25.

03:00 Plus one over the second resistor plus one over the third resistor.

03:06 So doing this a little bit, we have, see, common denominator does do 50.

03:14 So we'll have 2 over 50 plus 10 over 50 plus 5 over 50.

03:23 So continuing, we have 1 over the equivalent resistance is equal to 17 over 50, which therefore means, that are equivalent resistance, if we sort of invert these, is 50 over 17 oms.

03:40 So that's the equivalent resistance of these three in parallel.

03:44 Now let's add that to the final resistor in series, and we'll have our total resistance.

03:50 So our total resistance is 50 over 17 plus 10, which is approximately equal to, in decimal form, 12 .94 oms.

04:01 So we have officially filled in the first piece of information in this table here.

04:07 Our total resistance is 12 .94 oms.

04:15 All right.

04:16 So let's go through the rest of these and try and fill in this table here.

04:22 And then we'll have our answer.

04:24 Let me just give us some room here.

04:27 When i teach this in class, i always tell my students that in this table here, if they know two of the three in any row to just use omslaa to find a third.

04:39 So in this case, if i know the total voltage and the total resistance, i can find the total current using omsla.

04:45 So omslaah says v the voltage is equal to the current times the resistance.

04:50 In this case, i want to find the current, so i'm going to solve it for current, which is voltage divided by resistance.

04:55 So in this case, my total current coming out of the voltage or the battery is going to be 25 over 12 .94.

05:06 So that means the current exiting the battery is going to be 12 .94, 1 .1 .93 amps.

05:19 So my total current is 1 .93 amps.

05:23 All right.

05:27 That current, that total current is going to be the current that's going to r5.

05:31 Like i discussed earlier, because r5 is in series with the battery, whatever current is leaving the voltage source has to be reentering that voltage source.

05:42 So it doesn't matter which branch you go through, you always have to go through this one.

05:44 And so this one is also going to have a current of 1 .93 amps...

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