Digital Waveform Graph, Circuit Resistance, and Solenoids

So the answer would be C - the currents in R1 and R2 are twice the currents in R3 and R4. However, it is important to note that in a parallel circuit, the total current is equal to the sum of the currents through each branch. So in this case, the total current would be 1.2a, with 0.6a going through each resistor.8) Yes, you are correct. Poles 1 and 2 attract, making the answer D.For the wave question, you are on the right track. Refraction is the change in direction of a wave due to a change in velocity, which can
  • #1
Griffy11
27
0
Hi everyone,

20091013548196339100969917187507746.jpg


On this page, only questions 4 and 5 I need help for. For question 4, I really have no idea, and there's no real "working" that I can show to really say that I've given it a crack.

For question 5 on that page, I thought along these lines:

In series circuits, voltage changes.
In parallel circuits, current changes.
The first circuit is in series, with the second in parallel.
In the second circuit, I guess the current would be split equally across the two resistors, while in the first it would be the same in each resistor.
Therefore, I believe C is correct - however, I'm probably way off and it really needs clarification in my mind.

2009101355816339101028173437504329.jpg


For question 8 here, it's a bit like question 4 - I have no idea. Well - I think I do, but I know I'm wrong. Here's my line of thought:

Each "coil" is a solenoid, hence it has a magnetic field.
You determine the direction of the magnetic field using the right hand rule (not sure on even knowing how to use the right hand rule for this)
Then, that gives the direction of the magnetic field, so if two coils have opposite magnetic field directions, they will attract. If they have the same magnetic field direction, they'll repel.

So for example - pair 1.

Magnetic field moves to the right in both coils, thanks to the right hand rule. Therefore they repel? Is this the correct way to go about it?

Another question I had - A sound wave is observed to change its speed, direction of travel and wavelength, but not its frequency. What can we conclude from this?

A) The wave has been reflected
B) The wave has been refracted
C) The wave has been diffracted
D) The wave has been superimposed on another wave

I think the wave has been refracted, but I can't find any solid proof to back my answer up.

2. This question refers to the unlabelled graph below.

2009101432327633910874070468750248.jpg


Which pair of variables, when plotted against each other, would produce a graph like this?

A) The velocity of a water wave against its wavelength
B) The frequency of a sound wave against its wavelength
C) The amplitude of an electromagnetic wave against its frequency
D) The intensity of a light ray against its wavelength

I'm thinking B again, frequency against wavelength, as the frequency decreases when wavelength increases, but again - I'm not sure, although I'm more certain about this than the other one.

Thanks in advance everyone. :)
 
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  • #2
#4. "Digital" means on/off, 0 or 1, often either 5 Volts or 0 volts. Nothing in between.

#5. Let R be the resistance of one resistor. Use the formula relating voltage, current and resistance to find the current in each circuit. If you have trouble using "R", use R = 10 ohms and work it out as an example.

#8 "Each "coil" is a solenoid, hence it has a magnetic field.
You determine the direction of the magnetic field using the right hand rule (not sure on even knowing how to use the right hand rule for this)
Then, that gives the direction of the magnetic field,"
Good start! Draw the B field arrows on the diagram. Then mark N and S on the appropriate ends of each to show how it would look if it was a permanent magnet. Then you can use the fact that N and S attract, like poles repel.

Look up the characteristics of refraction. This is a good place to look if your textbook isn't handy:
http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html
Also Wikipedia.

The last question/graph is a classic inverse variation as in y = k/x.
Graph y = 12/x to make one for yourself.
Check the choices to see which one has that type of formula.
 
  • #3
Thanks a lot for the help Delphi! To go over what I've done now:

4) Since digital info is binary (1 or 0) I've concluded that only graph C is digital - as it only has two values - no gradual increases or decreases.

5) I'll try what you said. So, let R = 10 ohms, resistance of 1 resistor. Use V=IR.

For V=IR, we need I, so I = V/R
I=12/10
I=1.2a for the whole circuit.

In circuit 1, series, current = 1.2a through each resistor.
In circuit 2, parallel, current = 0.6a through each resistor.

So the answer would be C - the currents in R1 and R2 are twice the currents in R3 and R4. I have a feeling I've still done this wrong. In fact... I know I'm positive I've gone and done it wrong.

8) Glad to know I was on the right track for this one - thanks! To check:

Pair 1:
Both north poles are on the right of the solenoids = Attract.

Pair 2:
Left solenoid = left north pole. Right solenoid = right north pole. Attract.

Pair 3:
Left solenoid = right north pole. Right solenoid = left north pole. Repel.

Pair 4:
Left solenoid = right north pole. Right solenoid = left north pole. Repel

If that's correct, then poles 1 and 2 attract, making the answer D. Is this right?

For the wave question, I couldn't find any info on the link you gave me, but thanks for that link anyway - it'll be helpful for lots of other topics! :)

After a bit more research, the best thing I could find was that "refraction is the change in direction of a wave due to a change in velocity."

As for the unlabelled graph, thanks for the formula information. The formulas that I came up with for each option are:

A) v of water on wavelength, v=f * lambda, so f=v/lambda, which fits the y=k/x standard fine.
B) frequency on wavelength - v=f * lambda again, so doesn't fit.
C) amplitude on frequency - nothing to use here.
D) intensity on wavelength - again, no formula here.

I've probably overlooked something, but would that mean that A is the correct answer?

Thanks a lot for all your help! :D
 
  • #4
5) I'll try what you said. So, let R = 10 ohms, resistance of 1 resistor. Use V=IR.

For V=IR, we need I, so I = V/R
I=12/10
I=1.2a for the whole circuit.

In circuit 1, series, current = 1.2a through each resistor.
In circuit 2, parallel, current = 0.6a through each resistor.
In circuit 1, you have two 10 ohm resistors in series, so the resistance is NOT 10. In circuit 2, you have two 10 ohm resistors in parallel. Have to use 1/R = 1/10 + 1/10 to get their equivalent resistance.

Left solenoid = left north pole. Right solenoid = right north pole. Attract.
N...S and S...N looks like Repel to me.
Pair 4:
Left solenoid = right north pole. Right solenoid = left north pole. Repel
Check this one again - the right and solenoid. Hold your fingers curved as if going around one of the front wires, with thumb down. That makes the field to the right inside the loops - the equivalent of a permanent magnet with the N on the right.

Refraction. Say when light goes from air into a lens. The wave slows down. And bends if it hits the glass at an angle. The frequency must stay the same so that every time a crest hits the glass, it carries on in the glass.

For the graph, you were right the first time - frequency vs wavelength because it is f = v/w. You think f = v/w rather than v = f*w because that is what means graphing f vs w. y = k/x means graphing y vs x.
v = f*w would be used to graph v vs w.
 
  • #5
Thanks for the corrections on the solenoid Q8, in the end I worked out that pairs 1 and 4 attract, making the answer B. :)

Thanks again for the unlabeled graph question - looks like frequency/wavelength is it. :)

Also, was I correct in the digital waveform question? I assume since you didn't correct me on it then I was.

Lastly - the circuit. Circuit 1, series, two 10ohm resistors, so total resistance 20 ohms.
Circuit 2, two 10ohm resistors in parallel.
1/R = 1/10 + 1/10
1/R = 2/10
R = 5ohms - so the total resistance for circuit 2 is a quarter of the resistance in circuit 1... but that's not even an option. Looks like I messed it again. :S
 
  • #6
All correct, but that last bit about the resistors. One of the choices is for the parallel circuit to have 4 times as much current - that is the correct answer based on your 5 and 20 ohm resistances.
 
  • #7
Ahh of course, thanks a lot Delphi! One last question (which I posted above a while ago) and i'll stop bugging you. ;)

A sound wave is observed to change its speed, direction of travel, and wavelength, but not its frequency. What can we conclude from this?

A) The wave has been reflected
B) The wave has been refracted
C) The wave has been diffracted
D) The wave has been superimposed on another wave.

I think B - refracted, it seems obvious, but someone else got D, the wave was superimposed. I thought i'd just double check it.

Thanks! :)
 
  • #8
If it just got superimposed, I don't see why it would have to change its speed, direction of travel, or wavelength (or why frequency can't change). You got the right answer: B. All of those things happen when, for example, a light ray passes from air to water at an angle less than ninety degrees.
 

1. What is a digital waveform graph?

A digital waveform graph is a visual representation of a digital signal, showing the amplitude and timing of the signal over time. It is commonly used in digital systems to analyze and troubleshoot digital circuits.

2. How is circuit resistance measured?

Circuit resistance is measured using an ohmmeter, which measures the amount of resistance a circuit has to the flow of electrical current. Resistance is measured in ohms (Ω) and can also be calculated using Ohm's law (R = V/I).

3. What is a solenoid?

A solenoid is an electrical device that converts electrical energy into linear motion. It consists of a coil of wire wrapped around a core, and when an electrical current is passed through the coil, it creates a magnetic field that causes the core to move.

4. How are solenoids used in everyday life?

Solenoids are used in a variety of everyday devices, such as door locks, printers, and washing machines. They are also used in automotive applications, such as in solenoid valves for fuel injectors and transmission control.

5. What factors affect the performance of a digital waveform graph?

The performance of a digital waveform graph can be affected by factors such as the quality of the signal, noise interference, and the bandwidth of the oscilloscope being used to display the graph. It is also important to properly set the trigger level and time scale to accurately capture the signal.

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