Question

Difference between series and parallel combination of capacitors

Answer 1

Answer:

Explanation:A parallel circuit is one where the current splits into two pathways at a point. An example is:

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Here, the current in the loop splits into two.

Let's say that the battery produces a current of 'I' A. It splits into two branches i1 and i2. Here:

I=i1+i2

When this occurs, the circuits are said to be in parallel. This results in the same potential difference occuring across the two branches. So the potential difference across R1 is equal to the potential difference across R2.

The same idea is for Capacitors. When the potential difference across the plates is equal, they are in parallel. When the current through them is equal, they are in series.

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Here, in circuit A, the capacitors are in series, while in B, they are in parallel.

Many a time, capacitors are represented in the following manner:

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Here, the top circuit represents 3 capacitors in series. A, B, C, D are the plates of the capacitors. Capacitor AB is one, BC is one and CD is one.

The bottom circuit represents 3 capacitors in parallel. Since they're connected at the same 3 points, they have the same potential difference. The three capacitors are: Capacitor AB, capacitor A′B′ and capacitor A′′B′′. This notation is mostly used when different mediums are insterted into the capactitors - thus changing the ϵr value for each. Students wonder how to solve it further! It's easy - all of them are in parallel! :D

Hope I helped you understand!

Answer 2

Answer:

$\bigodot\mathtt \pink{\frac { 1 } { C } = \frac { 1 } { C _ { 1} } + \frac { 1 } { C _ { 2 } } + \frac { 1 } { C _ { 3 } } + \cdots + \frac { 1 } { C _ { n } }} \\ \bigodot\mathtt \pink{\frac { 1 } { C } = \frac { 1 } { C _ { 1} } + \frac { 1 } { C _ { 2 } } + \frac { 1 } { C _ { 3 } } + \cdots + \frac { 1 } { C _ { n } }}$