Courses
Courses for Kids
Free study material
Offline Centres
More

Three capacitors having capacitance $20\mu F$ ,$30\mu F$ and $40\mu F$ are connected in series with A $12V$ battery find the charge on each of the capacitors how much work has been done by the battery in charging the capacitors?

seo-qna
Last updated date: 29th Feb 2024
Total views: 21k
Views today: 0.21k
Rank Predictor
Answer
VerifiedVerified
21k+ views
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 $\therefore \dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}$.