Capacitors in Parallel and in Series MCQ Quiz in मराठी - Objective Question with Answer for Capacitors in Parallel and in Series - मोफत PDF डाउनलोड करा

CONCEPT:

Combination of capacitors:

Parallel combination	Series combination
When two or more capacitors are connected in such a way that their ends are connected at the same two points and have an equal potential difference for all capacitor is called a parallel combination of the capacitor.	When two or more capacitors are connected end to end and have the same electric charge on each is called the series combination of the capacitor.
Equivalent capacitance (Ceq) for parallel combination: Ceq = C1 + C2 + C3	Equivalent capacitance (Ceq) in series combination: \(\frac{1}{{{C_{eq}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\)

Where C1 is the capacitance of the first capacitor, C2 is the capacitance of the second capacitor and C3 is the capacitance of the third capacitor​

CALCULATION:

Given C1 = C2 = C3 = C = 4 μF and Ceq = 6 μF

When capacitors are connected in parallel:

⇒ Ceq = C + C + C = 4 + 4 + 4 = 12 μF

• When capacitors are connected in series:

\(\Rightarrow \frac{1}{C_{res}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}\)

\(\Rightarrow C_{res}=\frac{C}{3}\) = 4/3 = 1.33 μF

• When two capacitors are connected in series with the third capacitor in parallel:

• Resultant of the series arrangement,

\(\Rightarrow \frac{1}{C_{s}}=\frac{1}{C}+\frac{1}{C}\)

\(\Rightarrow C_{s}=\frac{C}{2}\)

Equivalent capacitance is given by:

\(\Rightarrow C_{eq}=\frac{C}{2}+C\) = 1.5 C = 1.5 × 4 = 6 μF

• Thus two of the capacitors will be in series and one of them will be in parallel with them.
• Hence option 4 is correct.

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
  • In a capacitor, two conducting plates are connected parallel to each other and separated by an insulating medium carrying charges of equal magnitudes and opposite signs.
  • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or a semi-conductor called a dielectric.​

​1. Capacitors in series:

• When two or more capacitors are connected one after another such that the same charge gets generated on all of them, then it is called capacitors in series.
• The net capacitance/equivalent capacitance (C) of capacitors in series is given by,

\(⇒\frac{1}{C} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}+...+ \frac{1}{{{C_n}}}\)

2. Capacitors in parallel:

• When the plates of two or more capacitors are connected at the same two points and the potential difference across them is equal, then it is called capacitors in parallel.
• The net capacitance/equivalent capacitance (C) of capacitors in parallel is given by,

\(⇒ C = C_1+ C_2+...+ C_n\)

CALCULATION:

The given diagram is,

         -----(1)

In figure 1 the equivalent capacitance between XY is given as,

⇒ C1 = C + C

⇒ C1 = 2C

In figure 1 the equivalent capacitance between YZ is given as,

⇒ C2 = C + C + C

⇒ C2 = 3C

Figure 1 can be drawn as,

        -----(2)

From figure 2 the equivalent capacitance between AB is given as,

\(⇒\frac{1}{C_{AB}} = \frac{1}{{{C}}}+\frac{1}{{{2C}}}+\frac{1}{{{3C}}}+\frac{1}{{{C}}}\)

\(\Rightarrow C_{AB}=\frac{6C}{17}\)

• Hence, option 2 is correct.

CONCEPT:

• Capacitance: The capacitance tells that for a given voltage how much charge the device can store.

Q = CV

where Q is the charge in the capacitor, V is the voltage across the capacitor and C is the capacitance of it.

• In the Parallel circuit, the equivalent capacitance is the algebraic sum of all the capacitance.
• And in the Series circuit, the reciprocal of the equivalent capacitance is the algebraic sum of all the reciprocal of the capacitance.

Ceq = C1 + C2 + C3 +......  (In parallel)

1/Ceq = 1/C1 + 1/C2 + 1/C3 +...... (in series)

CALCULATION:

Given that three equal capacitors in series provide 1µF capacitance

Let all the capacitors are C.

So their effective capacitance when connected in series

1/Ceq = 1/C1 + 1/C2 + 1/C3 

1/Ceq = 1/C + 1/C + 1/C

1/Ceq = 3/C

C = 3C_eq = 3 × 1µF = 3 µF

Each capacitors have capacity of 3 µF.

effective capacitance when connected in parallel

Ceq = C1 + C2 + C3 

Ceq = C + C + C

Ceq = 3C = 9 µF

So the correct answer is option 4.

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
  • In a capacitor, two conducting plates are connected parallel to each other and separated by an insulating medium carrying charges of equal magnitudes and opposite signs.
  • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or a semi-conductor called a dielectric.​

​1. Capacitors in series

• When two or more capacitors are connected one after another such that the same charge gets generated on all of them, then it is called capacitors in series.
• The net capacitance/equivalent capacitance (C) of capacitors in series is given by,

\(⇒\frac{1}{C} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}+...+ \frac{1}{{{C_n}}}\)

2. Capacitors in parallel

• When the plates of two or more capacitors are connected at the same two points and the potential difference across them is equal, then it is called capacitors in parallel.
• The net capacitance/equivalent capacitance (C) of capacitors in parallel is given by,

\(⇒ C = C_1+ C_2+...+ C_n\)

CALCULATION:

The given diagram is,

    -----(1)

Figure 1 can be drawn as,

     -----(2)

In figure 2 the equivalent capacitance in the upper and lower branch of AB is given as,

\(⇒\frac{1}{C_1} = \frac{1}{{{4μ F}}} + \frac{1}{{{4μ F}}}\)

⇒ C1 = 2μF

Figure 2 can be drawn as,

      -----(3)

So the equivalent capacitance between AB is given as,

⇒ C = C₁ + C₁

⇒ C = 2μF + 2μF

⇒ C = 4μF

• Hence, option 1 is correct.

CONCEPT:

• For a parallel plate capacitor of capacitance C placed in a potential difference V,
• The charge stored Q = CV
• The capacitance C = ϵ₀A/d (for air medium) and C' = kϵ0A/d (for a dielectric medium with dielectric constant k) ,(ϵ0 = permittivity of free space, A = area of the parallel plates, d = distance between the parallel plates)

EXPLANATION:

When the capacitor is fully charged with a battery with voltage V (say)

then the charge stored Q = CV (C= capacitance)

After that when the battery is removed the charge will be stored in the capacitor.

The charge in the capacitor plate Q remains unchanged.

Now, when a dielectric is inserted into the capacitor

The new capacitance C' = kϵ0A/d (dielectric constant k)

∴ C' = kC ⇒ So, the capacitance increases. (as k>1)

and as the voltage is removed the capacitor will discharge and

cause a charge flow which will increase the charges in the dielectric.

⇒ So, the charges on the dielectric surface will also increase. 

Hence the correct answer is option 3.

CONCEPT:

• The device that stores electrical energy in an electric field is called a capacitor.
• The capacity of a capacitor to store electric charge is called capacitance.
• When combination, where two or more capacitors are connected in such a way that their ends are connected at the same two points and have an equal potential difference across them, is called the parallel combination of a capacitor.
• Equivalent capacitance (Ceq) for parallel combination:

Ceq = C1 + C2

• When two or more capacitors are connected end to end and have the same electric charge on each is called a series combination of the capacitor.
• Equivalent capacitance (Ceq) in series combination:

\(\frac{1}{{{C}_{eq}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\)

Where C1 and C2 are two capacitors in the circuit.

CALCULATION:

• Here C₂, C_3, and C₄, are connected in series  therefore the equivalent capacitance is 

\(⇒ \frac{1}{{{C}_{eq}}}=\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+\frac{1}{C_4}\)

\(⇒ \frac{1}{{{C}_{eq}}}=\frac{1}{{{10}}}+\frac{1}{{10}}+\frac{1}{10}=\frac{3}{10}\)

⇒ C_eq = 10/3 μF

• Now, C_eq and C1 are connected in parallel, therefore the equivalent capacitance is 

⇒ Cpara = C₁ + Ceq 

\(\Rightarrow C_{para}=10 +\frac{10}{3}=\frac{40}{3}\, \mu F\)

The correct answer is option 1) i.e. 2C

CONCEPT:​

• Capacitor: A capacitor is an electrical component with two terminals used to store charge in the form of an electrostatic field in it. 
  • It consists of two parallel plates each possessing equal and opposite charges, separated by a dielectric constant.
  • Capacitance is the ability of a capacitor to store charge in it. The capacitance C is related to the charge Q and voltage V across them as:

\(⇒ C =\frac{Q}{V} \)​

• Equivalent capacitance of capacitors -
  • Connected in series: When n capacitors C₁, C₂, C₃, ... C_n are connected in series, the net capacitance (C_s) is given by

\(⇒ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}+ \frac{1}{C_3} + ... \frac{1}{C_n}\)

• Connected in parallel: When n capacitors C1, C2, C3, ... Cn are connected in parallel, the net capacitance (Cp) is given by

⇒ C_p = C1 + C2  + C3 +...  Cn

CALCULATION:

• The given circuit can be simplified as follows:

• The capacitors in arms AB and EF are in series. Hence the equivalent capacitance in each of these arms 

\(⇒ \frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C} ⇒ C_s = C/2\)

• The circuit arms AB, CD, and EF are in parallel and therefore, the equivalent capacitance in parallel is

⇒ C_P= C/2 + C + C/2 = 2C

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
  • In a capacitor, two conducting plates are connected parallel to each other and separated by an insulating medium carrying charges of equal magnitudes and opposite signs.
  • The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or a semi-conductor called a dielectric.​​

Capacitance:

• The charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates,

​⇒ Q ∝ V

⇒ Q =  CV

Where C = capacitance

​1. Capacitors in series

• When two or more capacitors are connected one after another such that the same charge gets generated on all of them, then it is called capacitors in series.
• The net capacitance/equivalent capacitance (C) of capacitors in series is given by,

\(⇒\frac{1}{C} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}+...+ \frac{1}{{{C_n}}}\)

2. Capacitors in parallel

• When the plates of two or more capacitors are connected at the same two points and the potential difference across them is equal, then it is called capacitors in parallel.
• The net capacitance/equivalent capacitance (C) of capacitors in parallel is given by,

\(⇒ C = C_1+ C_2+...+ C_n\)

CALCULATION:

Given C₁ = C2 = 10 F and Q = 20 C

• In the given figure the two capacitors are connected parallel so the equivalent capacitance of the arrangement is given as,

⇒ C = C₁ + C₂

⇒ C = 10 + 10 = 20 F     -----(1)

We know that if the charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates, then

⇒ Q =  CV     -----(2)

So the V is given as,

\(⇒ V=\frac{Q}{C}\)

\(⇒ V=\frac{20}{20}\)

⇒ V = 1 V

•  Hence, option 2 is correct.

Given:

Capacitor network with values: 30 μF, 10 μF, 6 μF, 2 μF, and 5 μF.

Concept:

Capacitors in series and parallel combinations. Use equivalent capacitance formulas.

Formula Used:

For capacitors in series: ¹⁄_Ceq = ¹⁄_C1 + ¹⁄_C2

For capacitors in parallel: C_eq = C₁ + C₂

Calculation:

⇒ V_x = V_y, since ³⁰⁄₆ = ¹⁰⁄₂

⇒ C_eq = (^{30 × 10}⁄_{30 + 10}) + (^{6 × 2}⁄_{6 + 2})

⇒ C_eq = (³⁰⁰⁄₄₀) + (¹²⁄₈)

⇒ C_eq = 7.5 + 1.5

⇒ C_eq = 9 μF

∴ Equivalent capacitance is 9 μF.

Concept:

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

\(\rm C_{series}=\frac{C}{n}\)
 
When 'n' capacitors are connected in parallel, the equivalent capacitance is:

\(\rm C_{parallel} = nC\)

Calculation:

Given:

n = 10

\(\rm \frac{C_{series}}{C_{parallel}} \rm =\frac{\frac{C}{n}}{nC}=\frac{1}{n^2}=\frac{1}{100}\)