circuit

[little boy]

My link Is it correct to say that R1 and R2 are in series , R1 and R3 are in parallel? God bless the people answering me.

 

Is there a simpler equivalent circuit?

Edited August 24, 2011 by stupid boy

[J.C.MacSwell]

My link Is it correct to say that R1 and R2 are in series , R1 and R3 are in parallel? God bless the people answering me.

 

Is there a simpler equivalent circuit?

 

R1 and R3 are in parallel and together are in series with parallel R2 and R4.

[little boy]

R1 and R3 are in parallel and together are in series with parallel R2 and R4.

 

Then it is incorrect to say that R1 and R2 are in series?

[TonyMcC]

It might help to look at the circuit like this:-

[Pantaz]

Then it is incorrect to say that R1 and R2 are in series?

Yes.

[DrRocket]

Then it is incorrect to say that R1 and R2 are in series?

 

Two resistors are in series if the current that flows through one must also flow through tne other.

 

Two resistors are in parallel in voltage impressed across one is also impressed across the other.

[little boy]

It might help to look at the circuit like this:-

 

Thank, now I am clear

 

How does it deduce that V1=R1/(R1+R2) x V while R1 and R2 are not in series? (highlighted)

 

image link

[TonyMcC]

Thank, now I am clear

 

How does it deduce that V1=R1/(R1+R2) x V while R1 and R2 are not in series? (highlighted)

 

image link

The first thing to realise is that long after the switch is closed the current through the capacitors will cease to flow( they will have accepted all the charge they can). This means that after that time you can ignore the capacitors when you calculate the voltages across the resistors. When you look at it this way, the resistors are in series and the formula used is correct. This is the standard voltage divider formula. http://en.wikipedia.org/wiki/Voltage_divider

[little boy]

The first thing to realise is that long after the switch is closed the current through the capacitors will cease to flow( they will have accepted all the charge they can). This means that after that time you can ignore the capacitors when you calculate the voltages across the resistors. When you look at it this way, the resistors are in series and the formula used is correct. This is the standard voltage divider formula. http://en.wikipedia....Voltage_divider

 

what happen after the switch is closed? I am confused.

[TonyMcC]

what happen after the switch is closed? I am confused.

Look at it this way :-

When the switch is open you can consider the two parts as separate circuits connected to the 12V supply. f you make the calculations you should be able to satisfy yourself that the voltage at the junction of the two resistors is 8V and the voltage at the junction of the two capacitors is 4V.

In the instant you close the switch the two junctions are connected together. This means they must be at the same voltage.

Firstly, this voltage drops to 4V because the charge held by a capacitor cannot change instantaneously. As time carries on the voltage will rise exponentially to 8V as the charges on the capacitors change. Therefore after a period of time the circuit stabilises with the charges held by the capacitors having changed and the voltage at the junction being 8V.

Hope this helps.

Edited August 26, 2011 by TonyMcC