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Circuit Analysis

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Question 1 0:12
  • Question 2 2:16
  • Question 3 2:33
  • Question 4 2:42
  • Question 5 3:18
  • Question 6 5:51
  • Question 7 6:00

Transcription: Circuit Analysis

Hi everyone and welcome back to Educator.com. 0000

In this mini-lesson we are going to do page 1 of APlusPhysics worksheet on circuit analysis, so with that you will find the link down below. 0002

We will dive right in. 0011

Number 1 -- A 30 ohm resistor, a 3 ohm resistor, and an unknown resistor (r) and two ammeters (A1) and (A2) are connected as shown with a 12 volt source and ammeter (A2) reads a current of 5 A. 0014

Determine the equivalent resistance of the circuit. 0025

Well, the way I am going to solve something like this is I am going to go through and I am going to make a VIRP table.0028

What I am going to do is I am going to name my resistors, so let us call that (R1) and that (R2). 0034

The elements of my circuits are (R1), (R2), and we will have a row here for total, have a voltage, a current, a resistance, and a power. 0040

I do not think I am going to need the power in this one, but we will write it in there, just in case. 0052

And we will fill in what we already know about the circuit. 0059

We know (R1) is 3 ohms, we know our total potential difference is 12 volts, and we know the current through the entire circuit is 5 A. 0061

We can figure out once we know two things in any row, we can figure out the others, so resistance is going to be V/I or 12/5, which is going to be 2.4. 0074

Since this is a parallel circuit, we are going to have the same potential drop across both resistors, so we can fill that in -- 12 and 12 -- and current is I = V/R, so this will be 4 A. 0086

If we have 4 A going through (R1) -- well if we had 5 A here, we have 4 A going through (R1), we must have 1 A going through (R2). 0099

Now I can figure out that resistance of our unknown resistor, R = V/I or 12 ohms; so determine the equivalent resistance of the circuit. 0111

The equivalent resistance of the circuit we already found out was 2.4 ohms. 0120

Now, as we do this, we are going to answer the next couple of questions using that same table. 0130

Number 2 -- Calculate the current measured by ammeter (A1).0137

Well, the current through (A1) we said was going to be 4 A, right from our VIRP table already, so 4 A. 0141

Number 3 -- Calculate the resistance of the unknown resistor, also right from out table and that was 12 ohms. 0153

Moving on to question 4 -- A 9 volt battery is connected to a 4 ohm resistor and a 5 ohm resistor as shown in the diagram. 0162

What is the current through the 5 ohm resistor? 0170

Well, I could do this with a VIRP table, but I can probably do this one much more simply because the current is just going to be total voltage over total resistance and this is easy to determine because it gives me both resistors and they are in series. 0173

So that will be 9 volts/4 ohms + 5 ohms or 9 ohms or just 1 A so the correct answer there is Number 1. 0186

Number 5 -- An 18 ohm resistor and a 36 ohm resistor are connected in parallel with a 24 volt battery.0196

A single ammeter is placed in the circuit to read its total current. 0204

Draw a diagram of the circuit. 0208

For our diagram, we have a 24 volt battery and then we have a 18 ohm resistor and it is connected in parallel with that 36 ohm resistor. 0210

So that is 18 ohms, that one is 36 ohms, which we will call (R1) and (R2). 0229

Draw the diagram of the circuit -- Done. 0236

Well now that we have it here, let us go ahead and complete the VIRP table. 0239

(R1) and (R2) are our elements and a row here for total and we will fill in V-I-R-P.0243

Now let us put in what values we happen to know already. 0262

Our total voltage is 24 volts, (R1) is 18 ohms, and (R2) is 36 ohms. 0265

As I look at this, because it is a parallel circuit, it is easy to see that the voltage drop across (R1) and (R2) is also 24 volts, so I can fill those in and current then is going to be V/R or 24/18, which is going to be 4/3 or 1.33 amps. 0273

The current through our two, I = V/R or 24/36 is 2/3 or we will round that off to 0.67 amps and that means that our total current -- if we have through (R1) we have 1.33 amps and through (R2) we have 0.67 amps -- those must come together to give us 2 amps as our total current. 0291

We can figure out now our total resistance, R = V/I, which will be 12 ohms, which makes sense. 0313

Our equivalent resistor should be less than any single resistor. 0321

While we are here, we could calculate the powers too. 0325

Power equals V × I or 24 × 4/3 = 32 W and 24 × 2/3 = 16 W and 24 × 2 = 48 W or we could have just added up the powers through each resistor. 0327

Now that we have this VIRP table, we can answer any of the other questions that are going to come up. 0343

For example, as we go through the next one -- calculate the equivalent resistance of the circuit. 0349

We just did that and the equivalent resistance was 12 ohms. 0355

Calculate the total power dissipated in the circuit. Our total power dissipated was 48 W. 0361

As we do these then, there is our 48 W, so...0370

...we have answered all seven questions on the worksheet. 0379

If you struggled with any of these, now would be a great time to go back and review the video lesson on circuit analysis and VIRP tables or if they went great -- Terrific -- time to move on, keep moving forward. 0382

Thanks so much for your time everyone and make it a great day!0394