Question

Three capacitors C1, C2 and C3 are connected in series. Obtain the relation for equivalent capacitance.
Answer.
Series Combination of Capacitors
To obtain small capacity from capacitors of large capacity, series combination of capacitors is adopted. In following diagram three capacitors of capacitances C1, C2 and C3 are joined in series. In this combination the charge stored on each capacitor will be the same as that given to combination. Since q is same for all capacitor, then their individual potential differences will be different say V1, V2, V3 respectively. If V be the total potential difference of the combination then

Answer 1

Explanation:

Charge on all the capacitor in series is same

But the potential on the equivalent capacitor will be equal to the sum of the potential of the individual capacitor connected in series

Q=C1V1=C2V2=C3V3(for each capacitor)

V1=Q/C1

V2=Q/C2

V3=Q/C3

V(total)=V1+V2+V3=Q/C(equivalent)

Q/C(equivalent)=Q/C1+ Q/C2+ Q/C3

1/C(equivalent)=1/C1+ 1/C2+1/C3

$C = \frac{C1×C2+C2×C3+C3×C1}{C1+C2+C3}$

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Answer 2

The equivalent capacitance is $\dfrac{C_{1}C_{2}C_{3}}{C_{2}C_{3}+C_{1}C_{3}+C_{1}C_{2}}$

Explanation:

Given that,

Three capacitors  C₁, C₂ and C₃ are connected in series.

We need to calculate the equivalent capacitance

Using formula of series

$\dfrac{1}{C_{eq}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}$

$\dfrac_{1}{C_{eq}}=\dfrac{C_{2}C_{3}+C_{1}C_{3}+C_{1}C_{2}}{C_{1}C_{2}C_{3}}$

$C_{eq}=\dfrac{C_{1}C_{2}C_{3}}{C_{2}C_{3}+C_{1}C_{3}+C_{1}C_{2}}$

Hence, The equivalent capacitance is $\dfrac{C_{1}C_{2}C_{3}}{C_{2}C_{3}+C_{1}C_{3}+C_{1}C_{2}}$

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Topic : capacitance

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