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

AI Thread Summary
The discussion revolves around finding the potential difference between nodes c and d in a circuit with a known potential difference of 100 volts between nodes a and b. Participants clarify that when capacitors are combined into a single equivalent capacitor, the original nodes c and d are effectively lost, as they are absorbed into the new configuration. It is emphasized that the equivalent capacitance calculated pertains to the entire circuit between a and b, not specifically between c and d. To determine the potential difference across c and d, one must consider the original series configuration of capacitors rather than simplifying them prematurely. Ultimately, the conversation concludes that understanding the placement of voltage sources is crucial for accurately determining potential differences in circuits.
  • #51
gracy said:
But it is derivation for resistance as we know ohm's law can't be applied for capacitors,how to derive this for capacitors?
Use Q=CV. In series circuit, Q is same on each capacitor. Can you proceed from here?
 
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  • #52
gracy said:
But it is derivation for resistance as we know ohm's law can't be applied for capacitors,how to derive this for capacitors?
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.
 
  • #53
gracy said:
Vo
What is Vo?
 
  • #54
gracy said:
What is Vo?
gracy said:
Vo = I × R2
Voltage across R2..
 
  • #55
In general?
 
  • #56
gracy said:
In general?
In general it is the output voltage..
 
  • #57
cnh1995 said:
it is the output voltage..
Voltage across last resistance/capacitor?
 
  • #58
gracy said:
Vs
Emf of battery?
 
  • #59
gracy said:
Voltage across last resistance/capacitor?
Not last. It can be across any component. That depends on the problem. Here in this capacitance problem, Vo is voltage between c and d.
 
  • #60
gracy said:
Emf of battery?
Right.
 
  • #61
How to recognize output voltage?I thought it is Voltage across last resistance/capacitor.
 
  • #62
gracy said:
How to recognize output voltage?I thought it is Voltage across last resistance/capacitor.
It is normally given in the problem itself.. However, the the terms input and output are commonly used in "systems" rather than in circuits like this. For example, power supply in our houses is Vin, the appliances connected to it are called 'load' and voltage across load is called output voltage.
gracy said:
Vo = I × R2
Here, R2 is the load.
 
  • #63
Did not understand.
 
  • #64
Gracy, what names you give to variables is not so important so long as you define them, are consistent in their use, and they convey some idea of what they are meant to represent. This just boils down to naming conventions for quantities. You can choose any variable names you want if you are so inclined, so long as you define them for your audience, usually by showing them on the circuit diagram.

It is typical to use ##V## for voltages, ##I## for currents, ##R## for resistances, and so on, and to give them subscripts or suffixes to distinguish them. Input or source voltages are often labelled ##V_i## or ##V_s## or ##V_o## (meaning "original" or initial value). What you want to take as an output voltage for whatever purposes you have at the time can be labelled any way you want, but common choices include ##V_{out}##, ##V_o## (Yes, it can be used for both input or output labels, just not both at the same time!). Some people prefer to use U instead of V for voltages. It's a cultural thing. It should not cause confusion because variables should ALWAYS be defined before they are used!

Sometimes you want to reference a voltage across a particular component and then you might find it convenient to use, for example ##V_{R1}## or ##V_{C3}## where R1 and R3 are labelled components in your circuit diagram.

It's up to you to define your variables and labels for a given problem. If you are writing equations using variables then you should know what they represent beforehand because you have defined them. If you want to call the potential across any component the "output" of your circuit, that's entirely up to you. There is no strict rule about where that component has to be located in the circuit diagram (although most prefer to read circuit diagrams from left to right, top to bottom, input towards output. It's just a convention, and not a strict rule).

Above you presented a derivation for the voltage across a particular resistor in a voltage divider. You used variables R1, R2, Rtot, I, Vs and Vo. Each was clear in its use and intent, and you defined mathematically that Vo = I × R2 and correctly presented the finished expression. Yet now you are coming back to ask what the variables mean, and how to recognize the output? This doesn't make sense: You defined and used them yourself.

You can extend the voltage divider from two to any number of resistors and choose the potential across anyone of them to be your "output".

Fig1.png
 
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  • #65
gracy said:
Did not understand.
Output voltage is the voltage across "expected" or "concerned" component. In your problem, C4 is the concerned component. It is specified in the problem.
 
  • #66
gneill said:
Yet now you are coming back to ask what the variables mean, and how to recognize the output? This doesn't make sense: You defined and used them yourself.
Actually I copied it .
 
  • #67
gracy said:
Actually I copied it .
Once you are done solving a number of problems, you'll be fluent in recognizing all that stuff and mainly, conventions.
 
  • #68
I want to understand what the formula means how to use it
php?image=https%3A%2F%2Fcdn.sparkfun.com%2Fassets%2Fe%2F7%2F6%2F3%2Fc%2F511968d9ce395f7c54000000.png

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.
 
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  • #69
gracy said:
I want to understand what the formula means how to use it
php?image=https%3A%2F%2Fcdn.sparkfun.com%2Fassets%2Fe%2F7%2F6%2F3%2Fc%2F511968d9ce395f7c54000000.png

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.
This formula is applicable only for resistors and not capacitors. I mean, you can't simply replace R1 with C1 and R2 with C2.
 
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  • #70
gracy said:
I want to understand what the formula means how to use it
php?image=https%3A%2F%2Fcdn.sparkfun.com%2Fassets%2Fe%2F7%2F6%2F3%2Fc%2F511968d9ce395f7c54000000.png

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.
Right. For resistors and inductors.
 
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  • #71
gracy said:
V in is always the emf of source
It can be source voltage but precisely, it is the total voltage across the series combination.
 
  • #72
gracy said:
I want to understand what the formula means how to use it
php?image=https%3A%2F%2Fcdn.sparkfun.com%2Fassets%2Fe%2F7%2F6%2F3%2Fc%2F511968d9ce395f7c54000000.png

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?
 
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  • #73
gneill said:
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
 
  • #74
gracy said:
Capacitors in Series all have the same current flowing through them
Right. But current ceases in steady state. What else is common?
 
  • #75
Please tell me am I correct in#68 that
Here in the following equation
%252F%252Fcdn.sparkfun.com%252Fassets%252Fe%252F7%252F6%252F3%252Fc%252F511968d9ce395f7c54000000.png

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.
 
  • #76
gracy said:
Please tell me am I correct in#68 that
Here in the following equation
%252F%252Fcdn.sparkfun.com%252Fassets%252Fe%252F7%252F6%252F3%252Fc%252F511968d9ce395f7c54000000.png

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.
 
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  • #77
gracy said:
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.)
 
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  • #78
cnh1995 said:
What else is common?
The equivalent charge is the same as the individual charges.
 
  • #79
gracy said:
The equivalent charge is the same as the individual charges.
Right.
 
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  • #80
Now how to proceed from here?
 
  • #81
gracy said:
Now how to proceed from here?
For how many capacitors in series? The formula will change as the number goes up.
 
  • #82
cnh1995 said:
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
 
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  • #83
gneill said:
Find out what is common to every component
charge
 
  • #84
gracy said:
Now how to proceed from here?
Calculate total charge from the equation Qtotal=Ceq*V. Then simply use Vconcerned=Qtotal/Cconcerned.
 
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  • #85
cnh1995 said:
Qtotal=Ceq*V.
100 multiplied by 8μC
800 μC=Qtotal
Right?
 
  • #86
gracy said:
100 multiplied by 8μC
800 μC=Qtotal
Right?
You have to consider only the series combination .
olu-png.92894.png
 
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  • #87
100 multiplied by 2μC
200 μC=Qtotal
Right?
 
  • #88
gracy said:
100 multiplied by 2μC
200 μC=Qtotal
Right?
Right. Qtotal for the series combination.
 
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  • #89
cnh1995 said:
Then simply use Vconcerned=Qtotal/Cconcerned.
Vcd=200 μC/6μF
=100/3 Volts
Right?
 
  • #90
gracy said:
Vcd=200 μC/6μF
=100/3 Volts
Right?
Correct:smile:
 
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  • #91
I also have to show charge distribution on capacitor plates.
 
  • #92
But we did not use voltage divider rule anywhere.
 
  • #93
gracy said:
But we did not use voltage divider rule anywhere.
We didn't have to. We simply followed the instructions from gneill in #82.
cnh1995 said:
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.
 
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  • #94
cnh1995 said:
We didn't have to
But I want to learn,I will create a separate thread on it.
cnh1995 said:
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 ?
 
  • #95
gracy said:
I also have to show charge distribution on capacitor plates.
You can do that using the information you calculated (the Q on each capacitor).
gracy said:
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.
 
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  • #96
gracy said:
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.
 
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  • #97
Ok.Then I will not learn voltage divider rule .
 
  • #98
  • #99
Okay. A quick visual summary of what's behind the general voltage division principle for components in series:

Fig2.PNG


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)
 
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  • #100
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}##
 
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