Answer

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**Hint:**First we calculate the equivalent capacitance and find the net charge stored in capacitors.

And to find the work done by the battery we calculate how much the charge battery sends to positive plate from negative plate of battery

**Complete step by step solution:**

First we make the circuit diagram of this combination which is given below

Here ${C_1} = 20\mu F$, ${C_2} = 30\mu F$ and ${C_3} = 40\mu F$

Battery $V = 12$ volt

Assume equivalent capacitance of circuit is ${C_{eq}}$

In series connection equivalent capacitance formula given as $\dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}$

Apply this formula

$ \Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{20}} + \dfrac{1}{{30}} + \dfrac{1}{{40}}$

$ \Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{{6 + 4 + 3}}{{120}}$

$ \Rightarrow \dfrac{1}{{{C_{eq}}}} = \dfrac{{13}}{{120}}$

So the equivalent capacitance

$\therefore {C_{eq}} = \dfrac{{120}}{{13}}\mu F$

Now we can calculate charge stored in capacitor

We know $C = \dfrac{q}{V}$

So $q = CV$

So the stored charge

$ \Rightarrow q = \dfrac{{120}}{{13}} \times 12$

$ \Rightarrow q = \dfrac{{1440}}{{13}}\mu C$

In series connection every capacitor has same amount of charge so every capacitor have same amount of charge $\dfrac{{1440}}{{13}}\mu C$

Step 2

Work done by the battery

Battery does work to carry charge from negative plate of capacitor to the positive plate of capacitor

In this process the battery does work equal to $qV$

So the work done by the battery to carry $\dfrac{{1440}}{{13}}\mu C$ charge is

$q = \dfrac{{1440}}{{13}} \times {10^{ - 6}}C$

$ \Rightarrow W = qV$

$ \Rightarrow W = \dfrac{{1440}}{{13}} \times {10^{ - 6}} \times 12$ Joule

Solving this

$ \Rightarrow W = 1329.23 \times {10^{ - 6}}$ Joule

**Note:**We use here series combination formula for calculating equivalent capacitance let’s talk about this. If capacitors connected in series then the charge stored in every capacitor same but the voltage across every capacitor different let take three capacitor connected in series voltage across capacitors ${C_1},{C_2}$ and ${C_3}$ is ${V_1},{V_2}$ and ${V_3}$ respectively.

${V_1} = \dfrac{q}{{{C_1}}},{V_2} = \dfrac{q}{{{C_2}}},{V_3} = \dfrac{q}{{{C_3}}}$

Net voltage $V = {V_1} + {V_2} + {V_3}$

$ \Rightarrow V = \dfrac{q}{{{C_1}}} + \dfrac{q}{{{C_2}}} + \dfrac{q}{{{C_3}}}$

If equivalent capacitance is ${C_{eq}}$ then $V = \dfrac{q}{{{C_{eq}}}}$

$ \Rightarrow \dfrac{q}{{{C_{eq}}}} = \dfrac{q}{{{C_1}}} + \dfrac{q}{{{C_2}}} + \dfrac{q}{{{C_3}}}$

So we get $\therefore \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}$.

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