What is the Potential Difference Between c and d in This Circuit?

[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}$

 

[gracy]

Similarly here
$V_i$=$I$$R_i$=$V_{in}$$\frac{R_i}{R_1+R_2+...+R_n}$

 

[epenguin]

Read #40. I have added to the "Relevant equations". When you take these into account you can solve this problem in 10 seconds... unless you prefer to be a pedantic student, but I see no virtue in taking days you could use for other study or something else over a 10 s problem.

 

[ehild]

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

Yes, thank you . I corrected it. You see how easy it is to make mistakes!
Yes i denotes any random element.
You can also notice that in case all n components of the chain are equivalent, the voltage across one of them is V_in/N . That was epenguin suggested you in #40.

 

[gneill]

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

Good catch! That's down to me getting sloppy with the cut & paste to duplicate the expressions, forgetting to "touch up" those entries.

I've updated the picture.