Switch in series/parallel Watch

Roqia
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Hi all,

Can someone explain how switches work? If you have a switch in a series, what happens to the Resistence, VOltage, and Current if you turn it on or off in series and parallel?
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teachercol
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In general an open switch behaves like an infinite resistor and a closed switch like a wire.

An open switch in a series circuit means that no current will flow through it or the rest of the circuit.

An open switch in a parallel branch means no current in that branch but has no effect on other branches

A closed switch in series has no effect on anything else.

A closed switch on its own in a parallel branch bypasses the component in other branches and you can ignore those branches.
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uberteknik
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(Original post by Roqia)
Hi all,

Can someone explain how switches work? If you have a switch in a series, what happens to the Resistence, VOltage, and Current if you turn it on or off in series and parallel?
Think of a switch as a device for introducing a break in a conductor with the ability to reinstate the conductor at will.

An ideal switch will have no resistance when closed and infinite resistance when open.

In this way, when closed, it allows current to flow but adds NO additional resistance to the conduction path and therefore does not introduce any additional voltage drops.

When open, the switch breaks the conduction path and NO current will flow.

In practice however, switches introduce small resistances (0.003 ohm is typical) and depending on the construction will have a maximum current and voltage carrying capability. Small switches may only withstand perhaps a few hundred mA with 50V across the contacts. Others are designed to carry upwards of 30A at 250V (domestic circuit breakers) etc.

With switches in series, all of the switches in the series path need to be closed in order for the conduction path to be created. If any single switch is opened, then the conduction path is broken.

With switches in parallel, if any of the switches are closed, then the conduction path will be completed.

Therefore:

Series switches must all be closed in order to make the circuit.

Parallel switches must all be open in order to break the circuit
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Roqia
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Thanks a lot for your help guys, but I am still stuck on this one. Could you please help?
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Roqia
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(Original post by uberteknik)
Think of a switch as a device for introducing a break in a conductor with the ability to reinstate the conductor at will.

An ideal switch will have no resistance when closed and infinite resistance when open.

In this way, when closed, it allows current to flow but adds NO additional resistance to the conduction path and therefore does not introduce any additional voltage drops.

When open, the switch breaks the conduction path and NO current will flow.

In practice however, switches introduce small resistances (0.003 ohm is typical) and depending on the construction will have a maximum current and voltage carrying capability. Small switches may only withstand perhaps a few hundred mA with 50V across the contacts. Others are designed to carry upwards of 30A at 250V (domestic circuit breakers) etc.

With switches in series, all of the switches in the series path need to be closed in order for the conduction path to be created. If any single switch is opened, then the conduction path is broken.

With switches in parallel, if any of the switches are closed, then the conduction path will be completed.

Therefore:

Series switches must all be closed in order to make the circuit.

Parallel switches must all be open in order to break the circuit
Thanks for your great explanation!! Could you perhaps please help me solve the question above, since it's still unclear
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Stonebridge
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(Original post by Roqia)
Thanks a lot for your help guys, but I am still stuck on this one. Could you please help?
Maybe it would have been better if you had posted this question initially, saying what you don't understand, rather than asking people to spend time writing explanations which may or may not be helpful, and then saying that what they have written doesn't actually address your problem...
Just a thought, to ensure you get faster help next time.
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mphysical
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(Original post by Roqia)
Thanks a lot for your help guys, but I am still stuck on this one. Could you please help?
When the switch is open ALL the current must flow through Y, so there will be a corresponding large voltage drop across Y.

But closing the switch creates a path of lesser resistance, meaning hardly any voltage drop across Y, therefore larger voltage drops across Z and X.

Remember all voltage drops should total 12V
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uberteknik
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(Original post by Roqia)
Thanks for your great explanation!! Could you perhaps please help me solve the question above, since it's still unclear
Please note Stonebridge's comment for future posts!

How to solve?

With all things circuit wise, start with the information given and work through it like a crossword puzzle - filling in the gaps from the clues given as you progress.

It's a good idea to copy out the diagram and mark up the information as you go along.

Open switch:

Known Information.

a) All resistors are equal.

b) In a series circuit, the resistances in the current path sum.

You are told the current through the Y resistor is 20mA. Since the switch is open, NO current will flow via the switch. All of the series current through resistor Y (20mA) must also flow through resistors X and Z.

Deduction

That means 20mA must therefore flow out of the battery terminal and 20mA flows back into the opposite battery terminal.

All of the battery terminal voltage must be dropped across the series resistance in the path of the 20mA current.

c) You now have enough information to calculate the total series resistance.

d) Given that all resistances are equal, you can also calculate the resistance for each resistor, X,Y & Z.

Closed switch

Known information

The parallel switch now presents almost no resistance to current which then effectively 'bypasses' resistor Y. i.e. Resistor Y and the parallel switch can both be replaced by a conductor wire.

Deduction

The current between the battery terminals must therefore flow through the series combination of X and Z only.
Hence calculate the new series path resistance R(X + Z)

Since the value of each resistor was calculated in d). above, the total current flowing out of the battery terminals and hence through X and Z can now be calculated. (V= 12V, Rtotal = R(x + z))

This gives the answer for the first column in the table.

We now also have enough information to calculate the voltage dropped across resistor Z and hence the answer to the second column.

What did you make the answer?

With sufficient practice and familiarity of the rules, the answers can be worked out in your head within 60 seconds or less and definitely if you use a calculator.
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Roqia
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(Original post by uberteknik)
Please note Stonebridge's comment for future posts!

How to solve?

With all things circuit wise, start with the information given and work through it like a crossword puzzle - filling in the gaps from the clues given as you progress.

It's a good idea to copy out the diagram and mark up the information as you go along.

Open switch:

Known Information.

a) All resistors are equal.

b) In a series circuit, the resistances in the current path sum.

You are told the current through the Y resistor is 20mA. Since the switch is open, NO current will flow via the switch. All of the series current through resistor Y (20mA) must also flow through resistors X and Z.

Deduction

That means 20mA must therefore flow out of the battery terminal and 20mA flows back into the opposite battery terminal.

All of the battery terminal voltage must be dropped across the series resistance in the path of the 20mA current.

c) You now have enough information to calculate the total series resistance.

d) Given that all resistances are equal, you can also calculate the resistance for each resistor, X,Y & Z.

Closed switch

Known information

The parallel switch now presents almost no resistance to current which then effectively 'bypasses' resistor Y. i.e. Resistor Y and the parallel switch can both be replaced by a conductor wire.

Deduction

The current between the battery terminals must therefore flow through the series combination of X and Z only.

Since the value of each resistor was calculated in d). above, the total current flowing out of the battery terminals and hence through X and Z can now be calculated. (V= 12V, Rtotal = R(x + z)

This gives the answer for the first column in the table.

We now also have enough information to calculate the voltage dropped across resistor Z and hence the answer to the second column.

What did you make the answer?

With sufficient practice and familiarity of the rules, the answers can be worked out in your head within 60 seconds or less and definitely if you use a calculator.
GOT IT!!!! thanks a looot
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uberteknik
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(Original post by Roqia)
GOT IT!!!! thanks a looot
What answer did you get?
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Roqia
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(Original post by uberteknik)
What answer did you get?
So when the switch is open, only Y is working and the current through Y is .02 A. So total R is 12/0.02 gives 600 Ohm.

When the eswitch is closed, only Z and X are working, so the total R is being shared. That means that R x is 300 ohm and R z is 300 ohm.
It is series, so the V is being shared as well, which means that V x is 6 volt and V z is 6 volt.

The Current is Voltage/Resistance so 6/300 is 0.02 Ampere is 20 mA.

Actually very easy, but just the first steps are tricky.

Thanks!
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uberteknik
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(Original post by Roqia)
So when the switch is open, only Y is working and the current through Y is .02 A. So total R is 12/0.02 gives 600 Ohm.
Incorrect. All three resistors (X, Y and Z) are in circuit with the total resistance of 600 ohms.

(Original post by Roqia)
When the eswitch is closed, only Z and X are working
Correct

(Original post by Roqia)
, so the total R is being shared. That means that R x is 300 ohm and R z is 300 ohm.
Incorrect.

Because all three resistances are equal, the individual values of X, Y and Z are 600/3 = 200 ohms each.

Making the total (closed switch) series resistance 200+200 = 400 ohms.

(Original post by Roqia)
It is series, so the V is being shared as well, which means that V x is 6 volt and V z is 6 volt.
Correct

(Original post by Roqia)
The Current is Voltage/Resistance so 6/300 is 0.02 Ampere is 20 mA.
Incorrect

The current is not 20mA because the series path resistance with the switch closed is not 600 ohms.

(Original post by Roqia)
Actually very easy, but just the first steps are tricky.

Thanks!
And therefore easy to make errors!
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Roqia
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(Original post by uberteknik)
Incorrect. All three resistors (X, Y and Z) are in circuit with the total resistance of 600 ohms.



Correct



Incorrect.

Because all three resistances are equal, the individual values of X, Y and Z are 600/3 = 200 ohms each.

Making the total (closed switch) series resistance 200+200 = 400 ohms.



Correct



Incorrect

The current is not 20mA because the series path resistance with the switch closed is not 600 ohms.


And therefore easy to make errors!
Hii!

The answer should be D, I have checked it ...
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uberteknik
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(Original post by Roqia)
Hii!

The answer should be D, I have checked it ...
Think again. The answer is not D.

With the switch open circuit:

Total series resistance = Rx + Ry + Rz

V/R(series open) = 12V / 20mA = 600 ohms.

Individual resistances X, Y and Z = 600 / 3 = 200 ohms each.



Spoiler:
Show
With the switch closed, the total series path comprises Rx + Rz only.

Total series (closed switch) resistance is therefore 200 + 200 = 400 ohms.

Total series current (flowing through Rx and Rz)
V/R(closed) = 12 / 400 = 30mA

Voltage dropped across Rz =
I(closed) x Rz = 30x10-3 x 200 = 6V

Answer is F
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Roqia
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(Original post by uberteknik)
Think again. The answer is not D.

With the switch open circuit:

Total series resistance = Rx + Ry + Rz

V/R(series open) = 12V / 20mA = 600 ohms.

Individual resistances X, Y and Z = 600 / 3 = 200 ohms each.



Spoiler:
Show
With the switch closed, the total series path comprises Rx + Rz only.

Total series (closed switch) resistance is therefore 200 + 200 = 400 ohms.

Total series current (flowing through Rx and Rz)
V/R(closed) = 12 / 400 = 30mA

Voltage dropped across Rz =
I(closed) x Rz = 30x10-3 x 200 = 6V

Answer is F
Sorry for asking, but is it possible that a resistance dissapears from a circuit? Because I didn't know that was possible.
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uberteknik
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(Original post by Roqia)
Sorry for asking, but is it possible that a resistance dissapears from a circuit? Because I didn't know that was possible.
Go back and read my post #3 carefully.

You are right that resistances do not simply disappear. But when a very low resistance is placed in parallel with a much larger resistance (as in this question), for all intents and purposes, the higher resistance can be ignored.

For example:

Assign the switch a contact resistance of 0.003 ohms.

In the closed condition, the switch resistance is then placed in parallel with Ry.

The effective combined parallel resistance of the two is then

\frac{1}{R_{parallel}} = \frac{1}{R_{switch}} + \frac{1}{R_y}

\frac{1}{R_{parallel}} = \frac{1}{0.003} + \frac{1}{200}

\frac{1}{R_{parallel}} = 333.33 + 5 x 10^{-3}

R_{parallel} = 2.9999 x 10^{-3}  ohms

The total series resistance with the switch closed is in reality (200 + 2.9999x10-3 + 200) = 400.0029999 ohms

Meaning, the 200 ohm resistor Ry is effectively 'bypassed' by the switch.

NB. This rule only holds true when the resistances are a few orders of magnitude difference.
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ErniePicks
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(Original post by teachercol)
In general an open switch behaves like an infinite resistor and a closed switch like a wire.

An open switch in a series circuit means that no current will flow through it or the rest of the circuit.

An open switch in a parallel branch means no current in that branch but has no effect on other branches

A closed switch in series has no effect on anything else.

A closed switch on its own in a parallel branch bypasses the component in other branches and you can ignore those branches.
what happens if you ahve a question to find net resistance when switch is open but the circuit network only has resistors and no source so you cannot see which direction current is flowing in, how would i calculate net resistance?
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uberteknik
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(Original post by ErniePicks)
what happens if you ahve a question to find net resistance when switch is open but the circuit network only has resistors and no source so you cannot see which direction current is flowing in, how would i calculate net resistance?
Resistance does not depend on the direction of current.
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ErniePicks
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(Original post by uberteknik)
Resistance does not depend on the direction of current.
But then if the switch is open would you just calculate resistance as if there was just a wire there? I dont understand how you would calculate resistance is there is an open switchin the middle of a circuit.
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uberteknik
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(Original post by ErniePicks)
But then if the switch is open would you just calculate resistance as if there was just a wire there? I dont understand how you would calculate resistance is there is an open switch in the middle of a circuit.
If the switch is open, you can replace it with an infinite resistance. i.e. no current will flow through the switch and all the current will flow through the resistance parallel to the switch.

If the switch is closed, replace it with a resistor valued at 0 ohms. i.e. all the current will flow through the switch and bypass anything in parallel with it.

Try it using ohms laws for a simple series and parallel combination of one resistor and one switch. Calculate the total resistance, p.d. and current through each component.





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