Capacitor directly connected to a DC battery.

In summary, initially when the key is closed, the capacitor is discharged and has voltages of (+ve terminal,10V) and (-ve terminal,0V).
  • #1
vissh
82
0
HeyoO everyone (^_^) and Thanks for your time.

Homework Statement


This question is about the "intial" point of time i.e. as soon as the key is closed.
Before the current starts to flow which charges the capacitor till the voltage of capacitor is equal to voltage across the battery.

A capacitor,C is directly connected to a DC battery of (say) 10V.
If the plates of the capacitor are taken to be at 0V, then what would be potential of "negative" and "positive" terminal
of battery relative to the plates?


Homework Equations


Electrons likes to flow from "Low" to "High" potential.

Finally, the plate A will be "Positively" charged and plate B will be negatively charged.
That means Electrons flowed from A -> X -> Positive terminal -> Negative terminal -> Y -> B which created those charges on Plate A and B.

This implies that Initially (the time at which this question is to be answered) Plate A [ 0V ] was at a lower potential than the positive terminal of the battery... And... Plate B [ 0V] was at a higher potential than the negative terminal of the battery.

A battery is of 10V implies that the potenital difference between the +ve and -ve terminal is of 10V.



The Attempt at a Solution


I think that the +ve terminal will be at +5V and -ve terminal will be at -5V.
That is +"Half of the voltage" on positive terminal and -"half of voltage" on negative terminal.

[Q.01] Am i right (o_O) ??


[Q.02] Will other pairs of (+ve Terinal , -ve terminal) work out?? :
Like (+9V , -1V) or (+4V, -6V).


[Q.03] The pair (+13V , +3V) will not be correct right ??
 

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  • #2
Ok I see the switch is in red.

So the initial conditions are switch open and the capacitor discharged.

It's necessary to state that the following voltages are measured with respect to the -ve terminal of the battery...

The A terminal of the capacitor (and point X) is the same voltage as the +ve terminal as the battery eg 10V.

The B terminal of the capacitor (and point Y) is the same voltage as the A terminal (eg 10V) because the capacitor is discharged.
 
Last edited:
  • #3
The way to think of this is that at t=0 when the switch is closed/ON the B terminal of the capacitor will be "pulled down" from 10V to 0V.
 
  • #4
Consider this circuit.

The discharged condition is represented by SW1 closed and SW2 open.
The charged condition is represented by SW1 open and SW2 closed.
 

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  • #5



I would like to confirm that your understanding is correct. The potential difference between the positive and negative terminals of the battery is 10V. When the capacitor is connected directly to the battery, the initial potential of the positive terminal will be +10V and the initial potential of the negative terminal will be 0V. This is because the capacitor acts like an open circuit before it is charged.

As the current starts to flow, electrons will flow from the negative terminal of the battery to the negative plate of the capacitor, creating a negative charge on the plate. At the same time, electrons will flow from the positive plate of the capacitor to the positive terminal of the battery, creating a positive charge on the plate. This will continue until the potential across the capacitor is equal to the potential of the battery, in this case, 10V.

To answer your questions:

1. Yes, you are correct. The potential of the positive terminal will be +5V and the potential of the negative terminal will be -5V when the capacitor is charged to 10V.

2. Other pairs of (+ve terminal, -ve terminal) will not work out because the potential difference between the positive and negative terminals of the battery is fixed at 10V. So, if one terminal has a potential of +9V, the other terminal will have a potential of +1V to maintain the potential difference of 10V.

3. Yes, the pair (+13V, +3V) will not be correct because the potential difference between the positive and negative terminals of the battery is fixed at 10V. Both terminals cannot have a potential greater than 10V.

Hope this helps clarify your understanding. Keep up the good work!
 

What is a capacitor?

A capacitor is an electronic component that stores electric charge. It consists of two conductive plates separated by an insulating material, also known as a dielectric.

How does a capacitor work when connected to a DC battery?

When a capacitor is connected to a DC battery, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field between the plates, causing the capacitor to store energy in the form of electric charge.

What is the purpose of connecting a capacitor to a DC battery?

The purpose of connecting a capacitor to a DC battery is to store electric charge and release it when needed. This can be useful for filtering, timing, and smoothing electrical signals.

What factors affect the capacitance of a capacitor directly connected to a DC battery?

The capacitance of a capacitor directly connected to a DC battery is affected by the distance between the plates, the surface area of the plates, and the type of dielectric material used.

Can a capacitor connected to a DC battery explode?

In rare cases, a capacitor connected to a DC battery can explode due to excessive voltage or current. It is important to use capacitors with appropriate voltage and current ratings and to handle them with caution to prevent explosions.

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