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Capacitors and Capacitance Test - 1

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Capacitors and Capacitance Test - 1
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  • Question 1
    1 / -0.25
    The electric polarization is-
    Solution

    CONCEPT:

    • Dielectrics: The material which are very poor conductors of electric current are called dielectrics. They are basically insulators.

    • Electric polarization: When a dielectric material is placed in an electric field then positive and negative charges of the molecules of the material feel electrostatic forces in opposite direction and they are separated by some distance within the molecules and they comes to the opposite end of the materials as shown in the above figure. Thus the molecules becomes an induced electric dipole. This process is called electric polarization.

     

    EXPLANATION:

    • The electric polarization is a process of rearranging positive charges and negative charges inside an insulator which are induced by an electric field. So option 4 is correct.
    • It is a process, it is not an electric dipole. So option 3 is wrong.
    • Due to this process, an induced electric dipole is formed inside an insulator.
    • Here there is no separation of charges from any charged materials. So options 1 and 2 are wrong.
  • Question 2
    1 / -0.25

    A spherical drop of capacitance 1 μF is broken into eight drops of equal radius. Then, the capacitance of each small drop is ____

    Solution

    CONCEPT:

    • A capacitor is a  device where two conductors are separated by an insulating medium that is used to store electrical energy or electrical charge
    • The capacitance is defined as the ability to store charge or it is the number of charges stored per unit potential in a capacitor.
      • Capacitance is given by

    \(\Rightarrow C=\frac{Q}{V}\)

    Where Q = Charge and V = Potential difference

    • The potential on a charged sphere is given by

    \(\Rightarrow V = \frac{K Q}{R}\)

    Where  K = Dielectric constant Q  = Charge, R  = Radius

    CALCULATION:

    Let r = radius after splitting, R = Radius before splitting

    • The potential on a charged sphere is given by

    \(\Rightarrow V = \frac{K Q}{R}\)

    The above of the equation can be rewritten as

    \(\Rightarrow \frac{Q}{V} = \frac{R}{K}\)

    We know \(C=\frac{Q}{V}\),  then the above equation can be written as

    \(\Rightarrow C = \frac{R}{K}\)

    As it is given that, C = 1 μF, then

    \(\Rightarrow \frac{R}{K} = 1 \ \mu F\)

    After splitting,

    \(\Rightarrow 8\times (\frac{4}{3}\pi r^3)=\frac{4}{3}\pi R^3\)

    \(\Rightarrow r=\frac{R}{2}\)

    • The potential of the smallest drop is given by

    \(\Rightarrow V = \frac{r}{K}\)

    Substituting the value of r in the above equation

    \(\Rightarrow V = \frac{R}{2 K}\)

    The substituting the value of \(\frac{R}{K} = 1 \ \mu F\) in  the above equation

    \(\Rightarrow V = \frac{1}{2 }\mu F\)

    Hence, option  1 is the answer

  • Question 3
    1 / -0.25

    Find the equivalent capacitance between A and B:

    Solution

    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.
  • Question 4
    1 / -0.25
    The bad conductor of heat is also known as-
    Solution

    option(2)

    CONCEPT:

    • Insulators or bad conductors: Bad Conductors are those which does not allow heat to pass through them.
    • Some examples of bad conductor:
      • wood
      • glass
      • plastic
      • diamond


    EXPLANATION:

    • The insulator is also named a bad conductor. Hence option 2 is correct.

    Additional Information

    • Conductors are those materials that allow heat and electricity to pass because of the movement of random motion of an electron in it.
    • Insulators are those materials that do not allow heat and electricity to pass because of the low conductivity.
  • Question 5
    1 / -0.25
    If a small value of capacitance is connected in parallel with a large value, the combined capacitance will be:
    Solution

    Concept:​

    Capacitor: A capacitor is an electrical component with two terminals used to store charge in the form of an electrostatic field. 

    • 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, ... Cn are connected in series, the net capacitance (Cs) 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:

    ⇒ Cp = C+ C2 + C+... Cn

    Explanation:

    let C1 be the lower value capacitance and C2 be the higher value capacitance

    If C1 and C2 are connected in parallel then

    Net capacitance (Cp) = C1 + C2 

    Net capacitance Cp > C2

    Therefore, If a small value of capacitance is connected in parallel with a large value, the combined capacitance will be higher than the large value capacitor.

  • Question 6
    1 / -0.25
    The capacitance of a capacitor is _______ relative permittivity.
    Solution

    Parallel plate capacitor:

    1) Parallel plate capacitors are the type of capacitors that have an arrangement of electrodes and insulating material (dielectric).

    2) The two conducting plates act as electrodes. There is a dielectric between them. This acts as a separator for the plates. 

    3) A capacitor stores energy in a dielectric dipole in the form of an electric field.

    4) Capacitors store energy by holding apart pairs of opposite charges. 

    5) If a capacitor has very high capacitance, then a small difference in plate voltage will lead to a huge difference in the number of electrons (total charge Q) on the two plates.

    The relation between capacitance, voltage, and charge stored is

    Q = C ΔV

    The capacitance for a parallel plate capacitor is given by

    \(C = \frac{{A\epsilon}}{d}\)

    Where ϵ = ϵ0ϵr

    ϵ0 = Permittivity of vacuum

    ϵr = Relative permittivity

    A = Area of the plate

    d = Gap between the plate

    ∴ The capacitance of a capacitor is directly proportional to relative permittivity.

  • Question 7
    1 / -0.25
    A capacitor that stores energy of 0.5 J, and has capacitance of 1 μF, has potential difference of ______ across it.
    Solution

    Concept:

    Energy stored in capacitor:

    • capacitor is a device to store energy.
    • The process of charging up a capacitor involves the transferring of electric charges from one plate to another.
    • The work done in charging the capacitor is stored as its electrical potential energy.
    • The energy stored in the capacitor is

    \(U = \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}C{V^2} = \frac{1}{2}QV\)

    Where Q = charge stored on the capacitor, U = energy stored in the capacitor, C = capacitance of the capacitor and V = Electric potential difference

    Calculation:

    Given: U = 0.5 J, C = 1 μF

    \(U = \frac{1}{2}C{V^2}\)

    \(0.5 = \frac{1}{2} \times 1 \times {10^{ - 6}} \times {V^2}\)

    V2 = 106

    V = 1000 volt

  • Question 8
    1 / -0.25
    If a capacitor stores 0.12 C at 10 V, then its capacitance is
    Solution

    Concept:

    The charge stored by a capacitor with an applied voltage V is given by:

    Q = C × V

    C = Capacitance of the capacitor

    V = Voltage applied across the capacitor

    Q = Charge stored

    Calculation:

    Given Q = 0.12 C

    V = 10 V

    Putting on the respective values, we can write:

    0.12 C = C × 10

    \(C=\frac{0.12}{10}F\)

    C = 0.0012 F

  • Question 9
    1 / -0.25
    X number of identical capacitors are connected in series and then in parallel. Then the ratio of the capacitances in parallel and series arrangement is-
    Solution

    The correct answer is option 2) i.e. X2

    CONCEPT:​

    • Capacitor: A capacitor is an electrical component with two terminals used to store charge in the form of an electrostatic field. 
      • 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, ... Cn are connected in series, the net capacitance (Cs) 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:

    ⇒ Cp = C+ C2  + C+...  Cn

    EXPLANATION:

    • Let C be the capacitance of each capacitor.

    • When identical capacitors are connected in series, the equivalent capacitance, Cs \(⇒ \frac{1}{C_s} = \frac{1}{C} + \frac{1}{C}+ \frac{1}{C} + ... \frac{1}{C} =\frac{X}{C}\)

    Thus, the equivalent capacitance in series, Cs = C/X

    • When the same capacitors are connected in parallel, the equivalent capacitance, Cp = C+ C2  + C+...  Cn = XC

    Thus, the equivalent capacitance in parallel, Cp = XC

    Ratio = \(\frac{C_p}{C_s} = \frac{XC}{C/X} = X^2\)

  • Question 10
    1 / -0.25

    Consider the following statements regarding capacitance

    1. Voltage of capacitance cannot change suddenly

    2. Current through a capacitance cannot change suddenly.

    Tick the correct answer:

    Solution

    Concept:

    Capacitance:

    • Capacitance is defined as it is the capability of an element to store electric charge within it. It is also defined the amount of charge
      required to create a unit p.d. between its plates.
    • Capacitance is the property that possesses the capacitor. Its unit is Farad.
    • It opposes the sudden change in voltage.

     

    It is given by,

    \(C=\frac{Q}{V} Farad\)

    Inductance: 

    • Inductance is the property of a material by virtue of which it opposes any changes of magnitude or direction of current passing through the conductor.
    • Inductance is said to be one henry when the current through a coil of the conductor changes at the rate of one ampere per second induction one volt across the coil.
    • The unit of inductance is Henry
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