physics -circuits

[bemanos]

i have this circuit

a)while the battery is connected what is the voltage of the capacitor
b)how long after the battery has disconected will the capacitor have a voltage of 1/10 of the initial one?.

 

can you tell me how to work it out? for question a) i try to find the equivalent resistance ,by combining R1 R8 (in series) and R4 R2(in series). Can i then combine R1R8-R4R2(parallel)?Any ideas?

[studiot]

Do you know what a potential divider is?

 

 

Edit It says you are viewing this right now.

Why take so long to reply yes or no?

 

Then perhaps you can get some help.

 

I'm going for a cup of tea.

Edited July 1, 2013 by studiot

[bemanos]

anything to do with this equation ? I=E/R_total E, is the EMF

 

I=10/1+8+4+2=10/15 A is that right?

Edited July 1, 2013 by bemanos

[studiot]

OK

 

A potential divider is made up of two resistors in series with a potential (voltage) across them.

It is one of the most important and basic circuit building blocks.

 

I have drawn this in Fig1.

 

The total voltage is divided by the resistors so if we take the voltage at their junction - point A - the voltage is divided in the ratio given bythe formula.

 

In Fig 2

I have redrawn your bridge circuit as two potential dividers side by side. Can you see this?

I have left you to fill in the resistors values and calculate the voltages at points A and B.

 

Let us know how you get on

Edited July 1, 2013 by studiot

[bemanos]

ok thanks for the answer helped alot

[ewmon]

I think a principle to understand is that when a voltage first acts on a capacitor, it has very little resistance, just like a short circuit. If the voltage is DC, a voltage builds within the capacitor which, at time = infinity, will equal the voltage applied to the resistor, and the current will stop, just like an open circuit. So, whether the charged capacitor is in or out of the circuit, it does not change the circuit.

 

It seems that question "a" assumes that time = infinity, so if you remove the capacitor from the circuit, you might see better what studiot means by voltage dividers.

[studiot]

If you have understood the answer to the first part of the question and correctly calculated the voltage across the capacitor between A and B

 

Do you realise why this is the voltage across the capacitor?

 

You are in a position to tackle part 2 of the question.

 

In part 2 the battery has been removed and the capacitor now discharges through the resistors.

 

Can you see that the two series combinations of resistors forming the potential dividers in part 1 are now in parallel across the capacitor?

 

Can you calculate the equivalent resistance of this?

 

Once you have this equivalent resistance can you see how to use it to calculate the rate of discharge of the capacitor?