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# Find ${C}_{e}$.A. $\dfrac {28}{3}\mu F$B. $\dfrac {15}{2}\mu F$C. $15 \mu F$D. None

Last updated date: 13th Jun 2024
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Hint: To solve this problem, consider a different group of capacitors. Find the equivalent capacitance of every group. Add the capacitance of capacitors which are connected in parallel directly. To find the equivalent capacitance of capacitors in series, take the reciprocal of their capacitances and add them. Then take the reciprocal of that obtained capacitance. This will give the equivalent capacitance of every group. Finally combine these equivalent capacitances to get the equivalent capacitance of the circuit.

Let ${C}_{1} = 23 \mu F$
${C}_{2} = 2 \mu F$
${C}_{3} = 12 \mu F$
${C}_{4} = 13 \mu F$
${C}_{5} = 1 \mu F$
${C}_{6} = 10 \mu F$
${C}_{7} = 1 \mu F$
Capacitors ${C}_{1}$ and ${C}_{3}$ are connected in parallel, So, the equivalent capacitance across them is given by,
${{C}_{eq}}_{1} = {C}_{1} + {C}_{3}$
Substituting the values in above equation we get,
${{C}_{eq}}_{1}= 23 + 12$
$\Rightarrow{{C}_{eq}}_{1}=35 \mu F$
Capacitors ${C}_{4}$ and ${C}_{5}$ are connected in series, So, the equivalent capacitance across them is given by,
$\dfrac {1}{{{C}_{eq}}_{2}}=\dfrac {1}{{C}_{4}} + \dfrac {1}{{C}_{5}}$
Substituting the values in above equation we get,
$\dfrac {1}{{{C}_{eq}}_{2}}=\ \dfrac {1}{13} + \dfrac {1}{1}$
$\Rightarrow \dfrac {1}{{{C}_{eq}}_{2}}= \dfrac {1}{13} +1$
$\Rightarrow \dfrac {1}{{{C}_{eq}}_{2}}=\dfrac {14}{13}$
$\Rightarrow{{C}_{eq}}_{2}= \dfrac {13}{14} \mu F$
Capacitor ${C}_{2}$ is in parallel with the ${{C}_{eq}}_{2}$. Thus, the equivalent capacitance will be,
${{C}_{eq}}_{3}= {C}_{2} + {{C}_{eq}}_{1}$
Substituting the values we get,
${{C}_{eq}}_{3}= 2 + \dfrac {13}{14}$
$\Rightarrow {{C}_{eq}}_{3}= \dfrac {41}{14} \mu F$
Now, in the circuit given, we can observe that ${{C}_{eq}}_{1}$ and ${{C}_{eq}}_{3}$ are in series. So, their equivalent combination is given by,
$\dfrac {1}{{{C}_{eq}}_{5}}= \ dfrac {1}{{{C}_{eq}}_{1}} +\ dfrac {1}{{{C}_{eq}}_{3}}$
Substituting the values in above equation we get,
$\dfrac {1}{{{C}_{eq}}_{5}}= \dfrac {1}{35} + \dfrac {14}{13}$
Solving the above expression we get,
${{C}_{eq}}_{5} = \dfrac {455}{503} \mu F$
Capacitors ${C}_{6}$ and ${C}_{7}$ are connected in series, So, the equivalent capacitance across them is given by,
$\dfrac {1}{{{C}_{eq}}_{6}}=\dfrac {1}{{C}_{6}} + \ dfrac {1}{{C}_{7}}$
Substituting the values in above equation we get,
$\dfrac {1}{{{C}_{eq}}_{6}}=\ \dfrac {1}{10} + \dfrac {1}{1}$
$\Rightarrow \dfrac {1}{{{C}_{eq}}_{6}}= \dfrac {1}{10} +1$
$\Rightarrow \dfrac {1}{{{C}_{eq}}_{6}}=\dfrac {11}{10}$
$\Rightarrow{{C}_{eq}}_{6}= \dfrac {10}{11} \mu F$
Now, ${{C}_{eq}}_{5}$ is in parallel with the ${{C}_{eq}}_{6}$. Thus, the equivalent capacitance will be,
${{C}_{eq}}_{7}= {C}_{5} + {{C}_{eq}}_{6}$
Substituting the values we get,
${{C}_{eq}}_{7}= \dfrac {455}{503} + \dfrac {10}{11}$
$\Rightarrow {{C}_{eq}}_{7}= 3.83 \mu F$
The equivalent capacitance is $3.83 \mu F$.

So, the correct answer is “Option D”.

Note: Students must remember that when the capacitors are connected in series, the total capacitance is less than at least any one of the series capacitors individual capacitance. When capacitors are connected in parallel, the total capacitance is the sum of all the capacitors’ capacitances. Students should remember that the formula for total capacitance is not the same as that for total resistance. So , students should not get confused between the formula for capacitance and resistance in series and capacitance and resistance in parallel.