Series and Parallel Connection of Capacitance MCQ Quiz in বাংলা - Objective Question with Answer for Series and Parallel Connection of Capacitance - বিনামূল্যে ডাউনলোড করুন [PDF]

Capacitor in Series: 

Consider two capacitors of capacitance C1 and C2 connected in series across supply having impedance Z1 and Z2 respectively as shown.

Applying Voltage division rule to the circuit,

The voltage across C1 is given as,

\(V_{C1}=V\times \frac{Z_1}{Z_1+Z_2}\) .... (1)

The voltage across C2 is given as,

\(V_{C1}=V\times \frac{Z_2}{Z_1+Z_2}\) .... (2)

V = VC₁ + VC₂

or, V = IZ₁ + IZ₂

The impedance Z1 and Z2 can be written as,

\(Z_1=\frac{1}{ω C_1},Z_2=\frac{1}{ω C_2}\)

Now, total capacitive impedance (Z = 1/ωC) will be,

Z = Z₁ + Z₂

or, \(\frac{1}{\omega C}=\frac{1}{\omega C_1}+\frac{1}{\omega C_2}\)

or, \(\frac{1}{ C}=\frac{1}{ C_1}+\frac{1}{ C_2}\) .... (3)

Hence, from equation (3), it is clear that, when two capacitors are connected in series, their total value of capacitance gets reduced.

Concept:

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors' capacitances.

\({C_{eq}}(parallel) = {C_1} + {C_2} + {C_3} + \ldots + {C_4}\)

When capacitors are connected in series, the total capacitance is less than the least capacitance connected in series, i.e.

\(\frac{1}{{{C_{eq}(series)}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

Calculation:

The capacitance of each capacitor = 500 μF

Both the capacitors are connected in parallel.

Ceq = 500 + 500 = 1000 μF

Concept of the star to delta conversion for capacitor;

Circuit Diagram:

All formula for resistance to convert from star to delta and delta to star is reverse for capacitor connection.

Star to delta conversion:

\({C_{AB}} = \frac{{{C_A}{C_C}}}{{{C_A} + {C_B} + {C_C}}}\)

\({C_{BC}} = \frac{{{C_A}{C_B}}}{{{C_A} + {C_B} + {C_C}}}\)

\({C_{CA}} = \frac{{{C_C}{C_A}}}{{{C_A} + {C_B} + {C_C}}}\)

Delta to Star conversion:

\({C_A} = \frac{{{C_{AB}}{C_{BC}} + {C_{BC}}{C_{CA}} + {C_{AB}}{C_{CA}}}}{{{C_{BC}}}}\)

\({C_B} = \frac{{{C_{AB}}{C_{BC}} + {C_{BC}}{C_{CA}} + {C_{AB}}{C_{CA}}}}{{{C_{CA}}}}\)

\({C_C} = \frac{{{C_{AB}}{C_{BC}} + {C_{BC}}{C_{CA}} + {C_{AB}}{C_{CA}}}}{{{C_{AB}}}}\)

Calculation:

By using the star-delta conversion formula,

\({C_{AB}} = \frac{{{C_A}{C_C}}}{{{C_A} + {C_B} + {C_C}}}\)

All capacitors are of the same value so, 

CAB = CBC = C_CA= C²/3C

= C/3 = 24 × 10^-6/3 = 8 × 10^-6 Farad

= 8μF

Concept:

When capacitors are connected in series, equivalent capacitance is given by

\(\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \ldots + \frac{1}{{{C_n}}}\)

When capacitors are connected in parallel, equivalent capacitance is given by

\({C_{eq}} = {C_1} + {C_2} + \ldots + {C_n}\)

Calculation:

Given that 10 capacitors are connected in series

Value of each capacitance = 10 μF

\(\frac{1}{{{C_{eq}}}} = \frac{1}{10} + \frac{1}{10} + \ldots upto\;5\;times\)

\(\Rightarrow {C_{eq}} = \frac{10}{{5}} = 2\;\mu F\)

Concept:

Capacitors in Series and Parallel:

The equivalent capacitance (C_eq) of capacitors connected in parallel is the sum of their individual capacitances:

C_eq = C₁ + C₂ + ... + C_n

The equivalent capacitance of capacitors connected in series is given by:

1 / C_eq = 1 / C₁ + 1 / C₂ + ... + 1 / C_n

Calculation:

We need to achieve an equivalent capacitance of 1.6 μF using four 4 μF capacitors.

One way to do this is by connecting two capacitors in parallel and then connecting two such combinations in series.

When two capacitors are connected in parallel:

⇒ C_eq1 = C + C = 4 μF + 4 μF = 8 μF

Now, when two such combinations of 8 μF each are connected in series:

⇒ 1 / C_eq2 = 1 / C_eq1 + 1 / C_eq1

⇒ 1 / C_eq2 = 1 / 8 μF + 1 / 8 μF = 2 / 8 μF

⇒ C_eq2 = 8 μF / 2 = 4 μF

Hence, the correct combination of capacitors to achieve 1.6 μF is:

Two capacitors in parallel and two such combinations in series.

∴ The correct answer is: Two in parallel and two in series.

Concept

Series connection of capacitors:

When 'n' capacitors are connected in series, the equivalent capacitance is given by:

\({1\over C_{eq}}={1\over C_{1}}+{1\over C_{2}}........{1\over C_{n}}\)

If all capacitors are equal, then:

\(C_{eq}={C\over n}\)

Parallel connection of capacitors:

When 'n' capacitors are connected in parallel, the equivalent capacitance is given by:

\(C_{eq}=C_1+C_2.......C_n\)

If all capacitors are equal, then:

\(C_{eq}=nC\)

Calculation

C = 5 μf

n = 5

\(C_{eq}={5\over 5}=1\space \mu F\)

The circuit can be drawn as

\(\begin{array}{l} \Rightarrow {C_{eq}} = 1 + \left( {1||{C_{eq}}} \right)\\ = 1 + \frac{{{C_{eq}}}}{{1 + {C_{eq}}}} \end{array}\)  

\(\begin{array}{l} {C_{eq}}\left( {1 + {C_{eq}}} \right) = 1 + {C_{eq}} + {C_{eq}}\\ C_{eq}^2 + {C_{eq}} = 2{C_{eq}} + 1 \end{array}\)  

\(\begin{array}{l} C_{eq}^2 - {C_{eq}} - 1 = 0\\ {C_{eq}} = \frac{{1 \pm \sqrt {1 + 4} }}{2} \end{array}\)  

\(= \frac{{1 + \sqrt 5 }}{2}\) 

\(\rm \begin{array}{l} c||c = c + c = 2c\\ \rm \frac{{c \times 2c}}{{c + 2c}} = \frac{{2c}}{3}\therefore {c_{eq}} = c + \frac{{2c}}{3} = \frac{{5c}}{3}\\ \rm {c_{eq}} = \frac{5}{3} \times 3 = 5\mu F \end{array}\)