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# Capacitors Grouping, Series Capacitors Based Assignment Help

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Electrostatics - Capacitors Grouping, Series Capacitors Based

**Physics Assignment Help >> Electrostatics >> Capacitors Grouping**

Capacitors Grouping

In many electrical circuits, capacitors are to be grouped suitably to obtain the desired capacitance. Two most common modes of grouping of condensers are: series grouping and parallel grouping

Capacitors in series

The capacitors are said to be connected in series between two points when we can proceed from one point to the other only thought one path.

Let + Q units of charge be given to the left plate of C_{1}. By electrostatic induction a charge – Q appears on inner side of right plate C_{1} and + Q appears on outer side of this plate. The + Q units of charge flow to left plate of C_{2} and so on* thus each capacitor receives the same charge of magnitude Q. as their capacities are different, therefore potential difference across the three capacitors are different:

**V**_{1} = Q / C_{1} V_{2} = Q / C_{2} V_{3} = Q / C_{3}

If Ca is the total capacitance of the combination of a single capacitor that would acquire charge Q when potential difference across its terminals is V) then

V = Q /C

As total potential drop V across the combination is the sum of potential drops **V**_{1} V_{2} V_{3} across **C**_{1} C_{2} C_{3} respectively,

**V = V**_{1} + V_{2} + V_{3}

**∴ Q / C = Q / C**_{1} + Q / C_{2} + Q / C_{3} + Q (1 / C_{1} + 1 / C_{2} + 1 / C_{3})

Or 1 / C = 1 / C_{1} + 1 / C_{2} + 1 / C_{3} V

If Q is the total charge on the parallel net work, then as total charges is the sum of charges on all capacitors.

**Q = Q**_{1} + Q_{3} + Q_{3} (2)

Let Cp be the equivalent capacitance in parallel (capacity of a single capacitor which would acquire the same total charge Q under the same potential difference V).

Q = Cp V

Putting these values in (12) we get

**C**_{p} V = C_{1} V + C_{2} V + C_{3} V = (C_{1} + C_{2 }+ C_{3}) V

∴ C_{p} = C_{1} + C_{2} + C_{3}

In general when n capacitors are connected in parallel, then

**C**_{p} = ∑ C_{i}

Equivalent capacitance of any number of capacitors joined in parallel is equal to sum of the individual capacitances. Obviously capacitance in parallel combination increases. Here **C**_{p} > the largest of **C**_{1}, C_{2}, C_{3}….C_{n}

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**Physics Assignment Help >> Electrostatics >> Capacitors Grouping**

_{1}. By electrostatic induction a charge – Q appears on inner side of right plate C

_{1}and + Q appears on outer side of this plate. The + Q units of charge flow to left plate of C

_{2}and so on* thus each capacitor receives the same charge of magnitude Q. as their capacities are different, therefore potential difference across the three capacitors are different:

**V**

_{1}= Q / C_{1}V_{2}= Q / C_{2}V_{3}= Q / C_{3}

**V**

_{1}V_{2}V_{3}across

**C**

_{1}C_{2}C_{3}respectively,

**V = V**

_{1}+ V_{2}+ V_{3}

**∴ Q / C = Q / C**

Or 1 / C = 1 / C

_{1}+ Q / C_{2}+ Q / C_{3}+ Q (1 / C_{1}+ 1 / C_{2}+ 1 / C_{3})Or 1 / C = 1 / C

_{1}+ 1 / C_{2}+ 1 / C_{3}V**Q = Q**

_{1}+ Q_{3}+ Q_{3}(2)**C**

∴ C

_{p}V = C_{1}V + C_{2}V + C_{3}V = (C_{1}+ C_{2 }+ C_{3}) V∴ C

_{p}= C_{1}+ C_{2}+ C_{3}

**C**

_{p}= ∑ C_{i}

**C**the largest of

_{p}>**C**

_{1}, C_{2}, C_{3}….C_{n}

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