Calculate the total resistance in a series circuit made up of resistances of 3Ω, 4Ω, and 5Ω.

How do you change the currents in a circuit

mel-nik [20]

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Eddy Current Testing

Introduction
Basic Principles
History of ET
Present State of ET

The Physics
Properties of Electricity
Current Flow & Ohm's Law
Induction & Inductance
Self Inductance
Mutual Inductance
Circuits & Phase
Impedance
Depth & Current Density
Phase Lag

Instrumentation
Eddy Current Instruments
Resonant Circuits
Bridges
Impedance Plane
Display - Analog Meter

Probes (Coils)
Probes - Mode of Operation
Probes - Configuration
Probes - Shielding
Coil Design
Impedance Matching

Procedures Issues
Reference Standards
Signal Filtering

Applications
Surface Breaking Cracks
SBC using Sliding Probes
Tube Inspection
Conductivity
Heat Treat Verification
Thickness of Thin Mat'ls
Thickness of Coatings

Advanced Techniques
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Background Pulsed ET
Remote Field Tech.

Quizzes

Formulae& Tables
EC Standards & Methods
EC Material Properties
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Current Flow and Ohm's Law

Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current, voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.

I = V / R

Where:

I =

Electrical Current (Amperes)

V =

Voltage (Voltage)

R =

Resistance (Ohms)

Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.

The Java applet below allows the user to vary each of these three parameters in Ohm's Law and see the effect on the other two parameters. Values may be input into the dialog boxes, or the resistance and voltage may also be varied by moving the arrows in the applet. Current and voltage are shown as they would be displayed on an oscilloscope with the X-axis being time and the Y-axis being the amplitude of the current or voltage. Ohm's Law is valid for both direct current (DC) and alternating current (AC). Note that in AC circuits consisting of purely resistive elements, the current and voltage are always in phase with each other.

Exercise: Use the interactive applet below to investigate the relationship of the variables in Ohm's law. Vary the voltage in the circuit by clicking and dragging the head of the arrow, which is marked with the V. The resistance in the circuit can be increased by dragging the arrow head under the variable resister, which is marked R. Please note that the vertical scale of the oscilloscope screen automatically adjusts to reflect the value of the current.

See what happens to the voltage and current as the resistance in the circuit is increased. What happens if there is not enough resistance in a circuit? If the resistance is increased, what must happen in order to maintain the same level of current flow?

4 0

3 years ago

Anyone trynna join ahahah<br> https://gotalk.to/yessirrrr

Mkey [24]

Answer:

nooo9ooooooooooooooooo

6 0

3 years ago

Read 2 more answers

A.

Elanso [62]

Answer:

Well, their are two answers in their. It would be Ask their Parent for assistance in persuading and Ask for an opportunity to earn extra credit:)

Explanation:

3 0

3 years ago

Read 2 more answers

A weightlifter is attempting a biceps curl with a 200 n barbell. the moment arm of the barbell about the elbow joint is 40 cm. t

CaHeK987 [17]

Weight of the barbell W = 200 Ndistance of the joint is r = 40 cm = 0.4 mtorque created by the weight at the joint is τ = F*r = 200 N*0.4 m = 80 N.mat equilibrium condition , Στ = force*distance - 80 N.m = 0 F'*0.4 - 80 N.m = 0 F'*0.4 = 80 force F' = 200 N

4 0

3 years ago

A ship's anchor weighs 5000N. It's cable passes over a roller of negligible mass and is wound around a hollow cylindrical drum o

deff fn [24]

Hi! Great first step would be to understand the scenario (in my opinion). So two great ways would be to draw a picture or rephrase it. If something else works, do that! You just need to "see" the situation so that you can take some away from it.

Then I think a good next step is to conceptualize everything. Put everything into a context like a physics book would. The anchor is pulled 5000N downward - that's weight. The roller will act like a pulley, and we can ignore it's properties except that it's part of a pulley system (we can ignore stuff because it has "negligible" mass and no other details are given). And then we have the hollow cylindrical drum with one radius measurement given; so we can think of this as a made-up shape with mass - a cylindrical soda can without a top or bottom (but no thickness) and a 380kg mass. The anchor is drops 16m. It hints at energy. The energy that the drum gets is all do to this anchor pulling on the rope (which is really just a means of transferring force, since we neglect its mass and get no details).

Feel free to pause here to make sure you can get the scenario in your head.

So, we want to know something about the barrel as it's rolling. The rotation rate. How many turns per some time. But don't worry yet, we can find a way to work that in. Since the rope pulls and spins the drum, the drum is spun, and gets energy. One way to find the kinetic energy of the spinning drum uses the radius, mass, and rate of rotation. More on that soon.

And how does having some equation with the drum's kinetic energy, radius, mass, and rate of rotation help? Well, we can find all of those except our rate of rotation and solve for the rate of rotation. The energy is the only mystery, but that all comes from the dropping anchor. Can we find that energy? Yeah, there's a way to find the energy that gravity gives our anchor based on it's the force and how far that force moves it.

So, first for the anchor. Linear work is simple: $W=F d$
So you have your force and distance we associate with the anchor, so you have your work. We'll call that " $W_1$ " when we need it.

Next the drum's situation. Thanks to http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html, we have the equation for kinetic energy.
Generally, we have <em></em> $KE=\frac12I\omega^2$ , and we need the " $I$ ," which deals with rotational inertia. That is pretty much how hard it is to rotate the drum based only on the idea that your getting the mass to move (acceleration). That site refers to our hollow drum as a "hoop," and gives says that we can consider the rotational inertia to be $I=MR^2$ . Now that we know the rotational inertia, we can use good old mathematical substitution to get the kinetic energy to look like
$KE=\frac12MR^2\omega^2$
And we can rearrange that to get
$\omega=\sqrt{\frac{2KE}{MR^2}}=\sqrt{\frac{2KE}{M}}\cdot\frac1R$

Since the energy change from the anchor's fall is the energy change of the drum, this $KE$ is the " $W_1$ " from before. So
$\omega=\sqrt{\frac{2W_1}{M}}\cdot\frac1R=\sqrt{\frac{2\left(F d\right)}{M}}\cdot\frac1R$

Now everything's set up. It's a matter of checking my work, carefully using a calculator, and making sure the answer makes sense (ie. this should be a lot of energy - much more than 1 Joule). Also, follow up by making sure you can do it again, alone. And feel free to ask or lookup questions you need along the way if there are missing pieces in your understanding.

Good luck! :)

5 0

3 years ago