Equivalent capacitance problem

[cnh1995]

V in is always the emf of source

It can be source voltage but precisely, it is the total voltage across the series combination.

 

[gneill]

I want to understand what the formula means how to use it

at the left side there will be voltage across the numerator on right side.
V in is always the emf of source
and denominator on the right side would be sum of all the component(resistance or capacitor) present.
numerator on the right side is always the component (resistance or capacitor) across which we want to find the voltage i.e V present on left side.

Refer to the derivation that you copied. The derivation used the fact that all the resistors in a series string share the same current. Then Ohm's law can be used with that current to find the voltage across anyone of the resistors in the string. It happened to be R2 in this example.

So in the case of resistors, since the common factor for all the resistors in a string is the shared current you determined that current first. That's $I = V_{in}/R_{eff}$, where I choose this time to name the total resistance $R_{eff}$, or the effective resistance of the string. You could call it $R_{tot}$ if you want. Then Ohm's law tells you the voltage across any resistor in the string by multiplying the current by that resistor value.
$$V_{Rj} = V_{in} \frac{R_j}{R_{eff}}$$
Now, what is the common factor shared by all capacitors in a series string?

 

[gracy]

Now, what is the common factor shared by all capacitors in a series string?

Capacitors in Series all have the same current flowing through them

 

[cnh1995]

Capacitors in Series all have the same current flowing through them

Right. But current ceases in steady state. What else is common?

 

[gracy]

Please tell me am I correct in#68 that
Here in the following equation

on the left side there is voltage across the numerator on right side.
V in is always the emf of source
and denominator on the right side would be sum of all the component(resistance or capacitor) present.
numerator on the right side is always the component (resistance or capacitor) across which we want to find the voltage i.e V present on left side.

 

[cnh1995]

Please tell me am I correct in#68 that
Here in the following equation

on the left side there is voltage across the numerator on right side.
V in is always the emf of source
and denominator on the right side would be sum of all the component(resistance or capacitor) present.
numerator on the right side is always the component (resistance or capacitor) across which we want to find the voltage i.e V present on left side.

Correct for resistors and inductors and not capacitors, as I said in #70.

 

[SammyS]

Please tell me am I correct in#68 that...

See post #72.

That is what gneill was doing; in a way that might lead to better overall understanding on your part.

(Post # changed in Edit.)

 

[gracy]

What else is common?

The equivalent charge is the same as the individual charges.

 

[cnh1995]

The equivalent charge is the same as the individual charges.

Right.

 

[gracy]

Now how to proceed from here?

 

[cnh1995]

Now how to proceed from here?

For how many capacitors in series? The formula will change as the number goes up.

 

[gneill]

For how many capacitors in series? The formula will change as the number goes up.

This is why it's better to understand the concept rather than try to memorize individual formulas. The concept applies to any number components in series, and a correct formula can be written down every time from the principle involved.

- Find out what is common to every component
- Calculate that common thing
- Apply the common thing's value to the individual component that you're interest in

 

[gracy]

Find out what is common to every component

charge

 

[cnh1995]

Now how to proceed from here?

Calculate total charge from the equation Q_total=C_eq*V. Then simply use V_concerned=Q_total/C_concerned.

 

[gracy]

Qtotal=Ceq*V.

100 multiplied by 8μC
800 μC=Qtotal
Right?

 

[cnh1995]

100 multiplied by 8μC
800 μC=Qtotal
Right?

You have to consider only the series combination .

 

[gracy]

100 multiplied by 2μC
200 μC=Qtotal
Right?

 

[cnh1995]

100 multiplied by 2μC
200 μC=Qtotal
Right?

Right. Q_total for the series combination.

 

[gracy]

Then simply use Vconcerned=Qtotal/Cconcerned.

Vcd=200 μC/6μF
=100/3 Volts
Right?

 

[cnh1995]

Vcd=200 μC/6μF
=100/3 Volts
Right?

Correct

 

[gracy]

I also have to show charge distribution on capacitor plates.

 

[gracy]

But we did not use voltage divider rule anywhere.

 

[cnh1995]

But we did not use voltage divider rule anywhere.

We didn't have to. We simply followed the instructions from gneill in #82.

It will be not as straightforward as it is for resistors. Because resistors is series add directly while capacitors in series don't.. You can derive it but can't memorize it because it will change with number of capacitors.

 

[gracy]

We didn't have to

But I want to learn,I will create a separate thread on it.

We simply followed the instructions from gneill in #82.

Can we solve all the problems with these steps that can be solved with voltage divider rule ?

 

[gneill]

I also have to show charge distribution on capacitor plates.

You can do that using the information you calculated (the Q on each capacitor).

But we did not use voltage divider rule anywhere.

You used the general voltage divider principle though, which boils down to the voltage divider rule when only two components are involved. The general principle is more flexible and can be applied in more situations without having to memorize a bunch of different formulas.

If you try it for a "string" of two capacitors and keep track of the algebra you will derive the voltage divider rule for two capacitors.

 

[cnh1995]

Can we solve all the problems with these steps that can be solved with voltage divider rule ?

Yes. Actually voltage divider rule for resistors is nothing but these steps. You can verify.

 

[gracy]

Ok.Then I will not learn voltage divider rule .

 

[gracy]

Ok.Thanks @gneill and @cnh1995

 

[gneill]

Okay. A quick visual summary of what's behind the general voltage division principle for components in series:

Used with just two components the principle yields the Voltage Divider rule(s) for that component type:
$$V_{R1} = V_{in} \frac{R1}{R1 + R2}~~~~~~V_{R2} = V_{in} \frac{R2}{R1 + R2}$$
$$V_{C1} = V_{in} \frac{C2}{C1 + C2}~~~~~~V_{C2} = V_{in} \frac{C1}{C1 + C2}$$

Note the subtle difference in the rules for the resistor and capacitor (pay close attention to what's in the numerators of the fractions). You don't need to remember this quirk if you always start with the basic principle.

(edit: Updated the figure to fix a subscript mixup)

 

[gracy]

in the picture why $V_{Rn}$ =$I$$R_n$ and $V_{Cn}$=$\frac{Q}{C_n}$ is not shown

This is true if only two resistance and capacitor are present

$V_{R1}$ = $V_{in}$ $\frac{R1}{R1 + R2}$~~~~~~$V_{R2}$ =$V_{in}$ $\frac{R2}{R1 + R2}$

$V_{C1}$ = $V_{in}$ $\frac{C2}{C1 + C2}$~~~~~~$V_{C2}$= $V_{in}$ $\frac{C1}{C1 + C2}$

But let's take the example of picture here there are five resistance

Then $V_{R1}$
$V_{R1}$ = $V_{in}$ $\frac{R1}{R1 + R2+R_j+R_(n-1)+R_n}$

Right?

I am unable to write the same for capacitors

I'll give it a try

$V_{C1}$=$V_{in}$ $\frac{C2C_jC_{n-1}C_n+C_1C_jC_{n-1}Cn+C1C2C_n-1+Cn+C1C2CjCn+C1C2CjC_n-1}{C1C2CjC_{n-1}Cn}$

 

[cnh1995]

VR1V_{R1} = VinV_{in} R1R1+R2+Rj+R(n−1)+Rn\frac{R1}{R1 + R2+R_j+R_(n-1)+R_n}

Right?

Right.

I am unable to write the same for capacitors

That's because series capacitors don't add directly.

 

[cnh1995]

I'll give it a try

VC1V_{C1}=VinV_{in} C2CjCn−1Cn+C1CjCn−1Cn+C1C2Cn−1+Cn+C1C2CjCn+C1C2CjCn−1C1C2CjCn−1Cn

I believe the summation part in the numerator should be in the denominator. Also, you can verify this formula by putting n=2.

 

[ehild]

in the picture why $V_{Rn}$ =$I$$R_n$ and $V_{Cn}$=$\frac{Q}{C_n}$ is not shown

It is just typo. The last formulas were meant V_Rn=IR_n and V_Cn=Q/C_n
Keep in mind that the voltages add when the components are connected in series. The total voltage across the chain of resistors is
V_in = I (R1+R2+...+Rn),
and in case of series capacitors is
V_in=Q/(C1+C2+...+CN).

In case of resistors,
$I=\frac{V_{in}}{R_1+R_2+...R_n}$ and the voltage across the i-th element is $V_i=IR_i=V_{in}\frac{R_i}{R_1+R_2+...+R_n}$.

In case of capacitors,
$Q=\frac{V_{in}}{1/C_1+1/C_2+...+1/C_n}$ and the voltage across the i-th element is $V_i=Q/C_i=V_{in}\frac{1/C_i}{1/C_1+1/C_2+...+1/C_n}$.

 

[gracy]

i-th element

Any random element ?

 

[gracy]

Should not it be $V_i$=Q/$C_i$=$V_{in}$$\frac{1/C_i}{1/C_1+1/C_2+...+1/C_n}$