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Lecture Comments (1)

0 answers

Post by Jae Chang Lee on May 5, 2015

way to easy

Electrical Current

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 0:13
  • Question 2 0:42
  • Question 3 2:01
  • Question 4 3:02
  • Question 5 3:52
  • Question 6 4:15
  • Question 7 4:37
  • Question 8 4:59
  • Question 9 5:50

Transcription: Electrical Current

Hello and welcome back to Educator.com. 0000

In this mini-lesson, we are going to work through page 1 of the APlusPhysics worksheet on electric circuits and you can find the link to that down below the video. 0002

With that, let us dive right in.0011

Number 1 -- What is the current through a wire if 240 C of charge pass through the wire in two minutes? 0013

Well, current is going to be charge divided by time or 240 C/2 min (time) = 120 s, which is going to be just 2 A, so our correct answer here must be Number 2, 2 A. 0021

Question 2 -- A 1.5 volt triple A-cell supplies 750 mA of current through a flash light bulb for five minutes, while a 1.5 volt C-cell supplies 750 mA of current through the same flash light bulb for 20 minutes. 0041

Compared to the total charge transferred by the triple A-cell through the bulb, the total charge transferred by the C-cell through the bulb is...0058

Let us figure out the charge through each one. 0065

If I = Q/t, then Q = It, so we will start with the triple A-cell. 0068

(Q) for triple A is going to be (I) for the triple A, which is 0.75 A × 5 min. (time) = 300 s or 0.75 × 300 = 225 C. 0075

Now for the C-cell, same formula -- Q for the C-cell is going to be equal to I × t and we have the same current, 0.75 A, but this works for 20 minutes or 1200 seconds... 0092

...so that is going to be 900 C, so that is four times greater than the charge transferred by the triple A-cell, so the correct answer must be Number 4, four times as great. 0107

Number 3 -- The current traveling from the cathode to the screen in a television picture tube is 5 × 10-5) A. 0122

How many electrons strike the screen in 5 s? 0129

Well, let us figure out the charge first. 0133

Charge is current times time, so that will be 5 × 10-5 A × 5 s (time), which is 25 × 10-5 C. 0135

But we want to know electrons, so let us convert coulombs to elementary charges. 0151

We will multiply that by 1 elementary charge -- 1.6 × 10-19 C. 0156

When I divide there, I am going to come up with a charge equal to about 1.6 × 1015 elementary charges or 1.6 ×1015 electrons, so the correct answer there must be 3. 0164

Problem 4 -- Charge flowing at the rate of 2.5 × 1016 e/s is equal to a current of...? 0182

Let us first change those elementary charges into coulombs. 0191

We could say that current is charge over time, so that is going to be 2.5 × 1016 e/1s times -- we want elementary charges to go away and we want coulombs and 1 elementary charge is 1.6 × 10-19 C...0197

...so when I multiply that out, I am going to come up with about 0.004 A or 4 × 10-3 A, which is Answer Number 3. 0217

On to problem Number 5 -- The current through a light bulb is 2 A. 0231

How many coulombs of electric charge pass through the light bulb in one minute? 0236

Charge is current time's time, which will be 2 A × 1 min. = 60 s or 120 C -- Answer Number 3. 0241

On to Number 6 -- If 10 C of charge are transferred through an electric circuit in 5 s, then the current in the circuit is...? 0255

Current is charge divided by times, so that will be 10 C/5 s or just 2 A -- correct answer there is Number 2. 0264

Moving on to Number 7 -- A charge of 30 C passes through a 24 ohm resistor in 6 seconds. 0276

What is the current through the resistor? 0283

Current is charge divided by time again, so 30 C/6 s is just going to be 5 A, so the correct answer is Number 2. 0286

Number 8 -- The diagram below shows two resistors (r1) and (r2) connected in parallel in a circuit having 120 volt power source. 0298

Resistor (r1) develops 150 W and (r2) develops an unknown power. 0308

Ammeter (A) in the circuit reads 0.5 A. Find the amount of charge passing through resistor (r2) down there in 60 s. 0313

Well, the key here is if this ampere reads 0.5 A, then that must be the current through (r2), so we know current through (r2) and we also know the time we should be able to find charge. 0322

Q = It, which is 0.5 A × 60 s (time) or just 30 C for our answer. 0335

One more -- What is the current in a wire if 3.4 × 1019 electrons pass by a point in this wire every 60 s? 0349

Our current is going to be charge divided by time, which is 3.4 × 10-19 elementary charges every 60 s, but we need to convert those elementary charges to coulombs...0359

...and I know that 1 elementary charge is 1.6 × 10-19 C, so my elementary charges will make a ratio of 1. 0374

I am going to determine that I have a current then of right around 0.091 A or 9.1 × 10-2 A, so the correct answer there is Number 3. 0384

That completes page 1 of the worksheet. 0400

If this went great -- Excellent -- you are probably ready for AP level problems and if it did not go so well, now would be a great time to go back and review the lesson about electric currents. 0403

Thanks for your time everyone and make it a great day!0412