Maths-
General
Easy

Question

The integrating factor of the differential equation fraction numerator d y over denominator d x end fraction left parenthesis x log space x right parenthesis plus b equals 2 log space x is given by

  1. e to the power of x
  2. log x
  3. log space left parenthesis log space x right parenthesis
  4. x

Hint:

In this question we have to find the integrating factor of the given equation. For this we will bring the equation in the form of fraction numerator d y over denominator d x end fraction plus a y equals b and IF =e to the power of integral a space d x end exponent.

The correct answer is: log x


    fraction numerator d y over denominator d x end fraction left parenthesis x log space x right parenthesis plus y equals 2 log space x
d i v i d i n g space x log x comma
rightwards double arrow fraction numerator d y over denominator d x end fraction plus fraction numerator y over denominator x space log x end fraction equals 2 over x
I F equals e to the power of integral fraction numerator 1 over denominator x space log x end fraction d x end exponent equals e to the power of log left parenthesis log x right parenthesis end exponent equals log space x

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    Related Questions to study

    General
    physics-

    As shown in figure, if the point C is earthed and the point A is given a potential of 2000 V, then the potential at point B will be

    Equivalent capacitance between points B a n d blank C is
    C to the power of ´ end exponent equals blank fraction numerator 10 cross times 10 over denominator 10 plus 10 end fraction plus 10 equals 15 mu F
    Now equivalent capacitance between points A a n d blank C is
    C to the power of ´ ´ end exponent equals fraction numerator 5 cross times 15 over denominator 15 plus 5 end fraction equals fraction numerator 75 over denominator 20 end fraction mu F
    Charge on capacitor of capacity 5mu F is
    Q equals C V equals fraction numerator 75 over denominator 20 end fraction cross times 2000 equals 7500 mu C
    (Since, potential at the point C will be zero)
    Now, potential difference across capacitor of 5mu F is
    V subscript A end subscript minus V subscript B end subscript blank equals fraction numerator Q over denominator 5 mu F end fraction equals fraction numerator 7500 mu C over denominator 5 mu C end fraction=1500volt
    As,V subscript A end subscript equals 2000volt
    Hence, V subscript B end subscript equals 2000 minus 1500 equals 500 volt.

    As shown in figure, if the point C is earthed and the point A is given a potential of 2000 V, then the potential at point B will be

    physics-General
    Equivalent capacitance between points B a n d blank C is
    C to the power of ´ end exponent equals blank fraction numerator 10 cross times 10 over denominator 10 plus 10 end fraction plus 10 equals 15 mu F
    Now equivalent capacitance between points A a n d blank C is
    C to the power of ´ ´ end exponent equals fraction numerator 5 cross times 15 over denominator 15 plus 5 end fraction equals fraction numerator 75 over denominator 20 end fraction mu F
    Charge on capacitor of capacity 5mu F is
    Q equals C V equals fraction numerator 75 over denominator 20 end fraction cross times 2000 equals 7500 mu C
    (Since, potential at the point C will be zero)
    Now, potential difference across capacitor of 5mu F is
    V subscript A end subscript minus V subscript B end subscript blank equals fraction numerator Q over denominator 5 mu F end fraction equals fraction numerator 7500 mu C over denominator 5 mu C end fraction=1500volt
    As,V subscript A end subscript equals 2000volt
    Hence, V subscript B end subscript equals 2000 minus 1500 equals 500 volt.
    General
    physics-

    The equivalent capacitance between points A a n d B for the combination of capacitors shown in figure, where all capacitances are in microfarad is

    all the three capacitors are connected in parallel. Therefore, equivalent capacitance between points A a n d B is
    C subscript e q end subscript equals 3 plus 3 plus 3 equals 9 mu F.

    therefore C to the power of ´ ´ end exponent equals C subscript 2 end subscript plus C subscript 3 end subscript equals 4 mu F
    As C to the power of ´ ´ end exponent a n d blank C subscript 1 end subscriptare in series,
    fraction numerator 1 over denominator C to the power of ´ ´ ´ ´ end exponent end fraction equals fraction numerator 1 over denominator blank to the power of ´ ´ end exponent end fraction plus fraction numerator 1 over denominator C subscript 1 end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 4 end fraction
    rightwards double arrow blank C to the power of ´ ´ ´ ´ end exponent equals blank 2 blank mu F
    Similarly, C subscript 4 end subscript a n d blank C subscript 5 end subscript are in parallel
    C to the power of ´ ´ ´ ´ ´ ´ end exponent equals blank 6 plus 2 equals 8 blank mu F
    C to the power of ´ ´ ´ ´ ´ ´ end exponent a n d blank C subscript 6 end subscript are in series
    fraction numerator 1 over denominator C ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 1 over denominator C ´ ´ ´ ´ ´ ´ end fraction plus fraction numerator 1 over denominator C subscript 6 end subscript end fraction equals fraction numerator 1 over denominator 8 end fraction plus fraction numerator 1 over denominator 8 end fraction
    rightwards double arrow blank C to the power of ´ ´ ´ ´ ´ ´ end exponent equals blank equals 4 blank mu F
    Now, C to the power of ´ ´ ´ ´ ´ ´ end exponent a n d blank C ´ ´ ´ ´ are in parallel.
    therefore blank C equals 4 mu F plus 2 mu F equals 6 mu F

    The equivalent capacitance between points A a n d B for the combination of capacitors shown in figure, where all capacitances are in microfarad is

    physics-General
    all the three capacitors are connected in parallel. Therefore, equivalent capacitance between points A a n d B is
    C subscript e q end subscript equals 3 plus 3 plus 3 equals 9 mu F.

    therefore C to the power of ´ ´ end exponent equals C subscript 2 end subscript plus C subscript 3 end subscript equals 4 mu F
    As C to the power of ´ ´ end exponent a n d blank C subscript 1 end subscriptare in series,
    fraction numerator 1 over denominator C to the power of ´ ´ ´ ´ end exponent end fraction equals fraction numerator 1 over denominator blank to the power of ´ ´ end exponent end fraction plus fraction numerator 1 over denominator C subscript 1 end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 4 end fraction
    rightwards double arrow blank C to the power of ´ ´ ´ ´ end exponent equals blank 2 blank mu F
    Similarly, C subscript 4 end subscript a n d blank C subscript 5 end subscript are in parallel
    C to the power of ´ ´ ´ ´ ´ ´ end exponent equals blank 6 plus 2 equals 8 blank mu F
    C to the power of ´ ´ ´ ´ ´ ´ end exponent a n d blank C subscript 6 end subscript are in series
    fraction numerator 1 over denominator C ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 1 over denominator C ´ ´ ´ ´ ´ ´ end fraction plus fraction numerator 1 over denominator C subscript 6 end subscript end fraction equals fraction numerator 1 over denominator 8 end fraction plus fraction numerator 1 over denominator 8 end fraction
    rightwards double arrow blank C to the power of ´ ´ ´ ´ ´ ´ end exponent equals blank equals 4 blank mu F
    Now, C to the power of ´ ´ ´ ´ ´ ´ end exponent a n d blank C ´ ´ ´ ´ are in parallel.
    therefore blank C equals 4 mu F plus 2 mu F equals 6 mu F
    General
    physics-

    The effective capacitance between points A a n d blank B is

    The effective capacitance between points A a n d blank B is

    physics-General
    General
    physics-

    A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 cm and the overlapping area of the plates is 5 blank c m to the power of 2 end exponent. The capacity of the unit is

    The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors.
    The capacity of each capacitor,
    C equals blank fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction
    equals blank fraction numerator 8.854 cross times 10 to the power of negative 12 end exponent cross times 5 cross times 10 to the power of negative 4 end exponent over denominator 0.885 cross times 10 to the power of negative 2 end exponent end fraction equals 0.5 p F
    Hence, the capacity of 8 capacitors
    equals 8 C equals 8 cross times 0.5 equals 4 blank p F

    A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 cm and the overlapping area of the plates is 5 blank c m to the power of 2 end exponent. The capacity of the unit is

    physics-General
    The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors.
    The capacity of each capacitor,
    C equals blank fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction
    equals blank fraction numerator 8.854 cross times 10 to the power of negative 12 end exponent cross times 5 cross times 10 to the power of negative 4 end exponent over denominator 0.885 cross times 10 to the power of negative 2 end exponent end fraction equals 0.5 p F
    Hence, the capacity of 8 capacitors
    equals 8 C equals 8 cross times 0.5 equals 4 blank p F
    General
    physics-

    A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

    The capacitor with air as the dielectric has capacitance
    C subscript 1 end subscript equals fraction numerator epsilon subscript 0 end subscript over denominator d end fraction open parentheses fraction numerator 3 A over denominator 4 end fraction close parentheses equals fraction numerator 3 epsilon subscript 0 end subscript A over denominator 4 d end fraction
    Similarly, the capacitor with K as the dielectric constant has capacitance C subscript 2 end subscript equals fraction numerator epsilon subscript 0 end subscript K over denominator d end fraction open parentheses fraction numerator A over denominator 4 end fraction close parentheses equals fraction numerator epsilon subscript 0 end subscript A K over denominator 4 d end fraction
    Since, C subscript 1 end subscript a n d blank C subscript 2 end subscript are in parallel
    C subscript n e t end subscript equals C subscript 1 end subscript plus C subscript 2 end subscript
    equals fraction numerator 3 epsilon subscript 0 end subscript A over denominator 4 d end fraction plus fraction numerator epsilon subscript 0 end subscript A K over denominator 4 d end fraction
    equals fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction open square brackets fraction numerator 3 over denominator 4 end fraction plus fraction numerator K over denominator 4 end fraction close square brackets
    equals fraction numerator C over denominator 4 end fraction left parenthesis K plus 3 right parenthesis

    A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

    physics-General
    The capacitor with air as the dielectric has capacitance
    C subscript 1 end subscript equals fraction numerator epsilon subscript 0 end subscript over denominator d end fraction open parentheses fraction numerator 3 A over denominator 4 end fraction close parentheses equals fraction numerator 3 epsilon subscript 0 end subscript A over denominator 4 d end fraction
    Similarly, the capacitor with K as the dielectric constant has capacitance C subscript 2 end subscript equals fraction numerator epsilon subscript 0 end subscript K over denominator d end fraction open parentheses fraction numerator A over denominator 4 end fraction close parentheses equals fraction numerator epsilon subscript 0 end subscript A K over denominator 4 d end fraction
    Since, C subscript 1 end subscript a n d blank C subscript 2 end subscript are in parallel
    C subscript n e t end subscript equals C subscript 1 end subscript plus C subscript 2 end subscript
    equals fraction numerator 3 epsilon subscript 0 end subscript A over denominator 4 d end fraction plus fraction numerator epsilon subscript 0 end subscript A K over denominator 4 d end fraction
    equals fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction open square brackets fraction numerator 3 over denominator 4 end fraction plus fraction numerator K over denominator 4 end fraction close square brackets
    equals fraction numerator C over denominator 4 end fraction left parenthesis K plus 3 right parenthesis
    General
    physics-

    Four capacitors are connected in a circuit as shown in the following figure. Calculate the effective capacitance between the points A a n d B.

    Effective capacitance of C subscript 2 end subscript a n d blank C subscript 3 end subscript
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 2 end fraction
    therefore blank C equals blank 1 mu blank F
    Now, C subscript 1 end subscript a n d blank C are in parallel, therefore effective capacitance C ´
    C to the power of ´ end exponent equals blank 1 plus 1 equals 2 mu blank F
    Now, C to the power of ´ end exponent a n d blank C subscript 4 end subscript are in series, therefore,
    effective capacitance between points A a n d blank B
    fraction numerator 1 over denominator C ´ ´ end fraction equals fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 3 over denominator 4 end fraction
    rightwards double arrow blank C to the power of ´ ´ end exponent equals fraction numerator 4 over denominator 3 end fraction mu F

    Four capacitors are connected in a circuit as shown in the following figure. Calculate the effective capacitance between the points A a n d B.

    physics-General
    Effective capacitance of C subscript 2 end subscript a n d blank C subscript 3 end subscript
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 2 end fraction
    therefore blank C equals blank 1 mu blank F
    Now, C subscript 1 end subscript a n d blank C are in parallel, therefore effective capacitance C ´
    C to the power of ´ end exponent equals blank 1 plus 1 equals 2 mu blank F
    Now, C to the power of ´ end exponent a n d blank C subscript 4 end subscript are in series, therefore,
    effective capacitance between points A a n d blank B
    fraction numerator 1 over denominator C ´ ´ end fraction equals fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 3 over denominator 4 end fraction
    rightwards double arrow blank C to the power of ´ ´ end exponent equals fraction numerator 4 over denominator 3 end fraction mu F
    General
    physics-

    Equivalent capacitance between A a n d B is


    The capacitors 2blank mu F and 2blank mu F of arm A C D are in series. So, their equivalent capacitance is 1blank mu F which is in parallel with capacitor of 1blank mu F of arm A D.
    So, equivalent capacitance now is 2blank mu F.
    This capacitance is now in series with 2blank mu F capacitance of arm B D which equivalents to 1blank mu F is in parallel with 1blank mu F capacitance of arm A B.
    So, final effective capacitance=2blank mu F.

    Equivalent capacitance between A a n d B is

    physics-General

    The capacitors 2blank mu F and 2blank mu F of arm A C D are in series. So, their equivalent capacitance is 1blank mu F which is in parallel with capacitor of 1blank mu F of arm A D.
    So, equivalent capacitance now is 2blank mu F.
    This capacitance is now in series with 2blank mu F capacitance of arm B D which equivalents to 1blank mu F is in parallel with 1blank mu F capacitance of arm A B.
    So, final effective capacitance=2blank mu F.
    General
    Physics-

    Effective capacitance between points A a n d blank B in the figure, shown is

    The points C a n d blank D will be at same potentials since,fraction numerator 3 over denominator 6 end fraction equals fraction numerator 4 over denominator 8 end fraction.
    Therefore, capacitance of 2mu F will be unaffected.
    So, the equivalent circuit can be shown as.

    The effective capacitance in upper arm in series, is given by
    C subscript 1 end subscript equals fraction numerator 3 cross times 6 over denominator 3 plus 6 end fraction equals fraction numerator 18 over denominator 9 end fraction equals 2 blank mu F
    The effective capacitance in lower arm in series is given by
    C subscript 2 end subscript equals blank fraction numerator 4 cross times 8 over denominator 4 plus 8 end fraction equals fraction numerator 32 over denominator 12 end fraction equals fraction numerator 8 over denominator 3 end fraction mu F
    Hence, the resultant capacitance in parallel is given by
    C equals C subscript 1 end subscript plus C subscript 2 end subscript equals 2 plus fraction numerator 8 over denominator 3 end fraction equals fraction numerator 14 over denominator 3 end fraction mu F

    Effective capacitance between points A a n d blank B in the figure, shown is

    Physics-General
    The points C a n d blank D will be at same potentials since,fraction numerator 3 over denominator 6 end fraction equals fraction numerator 4 over denominator 8 end fraction.
    Therefore, capacitance of 2mu F will be unaffected.
    So, the equivalent circuit can be shown as.

    The effective capacitance in upper arm in series, is given by
    C subscript 1 end subscript equals fraction numerator 3 cross times 6 over denominator 3 plus 6 end fraction equals fraction numerator 18 over denominator 9 end fraction equals 2 blank mu F
    The effective capacitance in lower arm in series is given by
    C subscript 2 end subscript equals blank fraction numerator 4 cross times 8 over denominator 4 plus 8 end fraction equals fraction numerator 32 over denominator 12 end fraction equals fraction numerator 8 over denominator 3 end fraction mu F
    Hence, the resultant capacitance in parallel is given by
    C equals C subscript 1 end subscript plus C subscript 2 end subscript equals 2 plus fraction numerator 8 over denominator 3 end fraction equals fraction numerator 14 over denominator 3 end fraction mu F
    General
    physics-

    The total energy stored in the condenser system shown in the figure will be

    6blank mu F and 3blank mu F capacitors are in series
    fraction numerator 1 over denominator C subscript 1 end subscript end fraction equals fraction numerator 1 over denominator 6 end fraction plus fraction numerator 1 over denominator 3 end fraction
    C subscript 1 end subscript equals 2
    C subscript 1 end subscript i s blank p a r a l l e l blank t o blank 2 blank mu F blank c a p a c i t o r
    therefore blank C subscript e q end subscript equals 2 plus 2 equals 4 mu F blank
    Total energy, U equals fraction numerator 1 over denominator 2 end fraction C V to the power of 2 end exponent
    equals fraction numerator 1 over denominator 2 end fraction cross times 4 cross times open parentheses 2 close parentheses to the power of 2 end exponent equals 8 mu J

    The total energy stored in the condenser system shown in the figure will be

    physics-General
    6blank mu F and 3blank mu F capacitors are in series
    fraction numerator 1 over denominator C subscript 1 end subscript end fraction equals fraction numerator 1 over denominator 6 end fraction plus fraction numerator 1 over denominator 3 end fraction
    C subscript 1 end subscript equals 2
    C subscript 1 end subscript i s blank p a r a l l e l blank t o blank 2 blank mu F blank c a p a c i t o r
    therefore blank C subscript e q end subscript equals 2 plus 2 equals 4 mu F blank
    Total energy, U equals fraction numerator 1 over denominator 2 end fraction C V to the power of 2 end exponent
    equals fraction numerator 1 over denominator 2 end fraction cross times 4 cross times open parentheses 2 close parentheses to the power of 2 end exponent equals 8 mu J
    General
    physics-

    In the given network, the value of C, so that an equivalent capacitance between A and B i s blank 3 mu F comma is

    3 equals fraction numerator fraction numerator 16 over denominator 5 end fraction C over denominator fraction numerator 16 over denominator 5 end fraction plus C end fraction
    Or C equals 48 mu F

    In the given network, the value of C, so that an equivalent capacitance between A and B i s blank 3 mu F comma is

    physics-General
    3 equals fraction numerator fraction numerator 16 over denominator 5 end fraction C over denominator fraction numerator 16 over denominator 5 end fraction plus C end fraction
    Or C equals 48 mu F
    General
    physics-

    A capacitor of capacitance 1 blank mu F is filled with two dielectrics of dielectric constant 4 and 6. What is the new capacitance?

    Initially, the capacitance of capacitor

    C equals fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction
    therefore blank fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction equals 1 mu F …(i)
    When it is filled with dielectric of dielectric constant K subscript 1 end subscript a n d blank K subscript 2 end subscript as shown, then there are two capacitors connected is parallel. So,
    C to the power of ´ ´ end exponent equals fraction numerator K subscript 1 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d end fraction plus fraction numerator K subscript 2 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d end fraction
    (as area becomes half)
    C ´ ´ blank equals fraction numerator 4 epsilon subscript 0 end subscript A over denominator 2 d end fraction plus fraction numerator 6 epsilon subscript 0 end subscript A over denominator 2 d end fraction equals fraction numerator 2 epsilon subscript 0 end subscript A over denominator d end fraction plus fraction numerator 3 epsilon subscript 0 end subscript A over denominator d end fraction
    Using Eq. (i), we obtain
    C to the power of ´ ´ end exponent equals blank 2 cross times 1 plus 3 cross times 1 equals 5 blank mu F

    A capacitor of capacitance 1 blank mu F is filled with two dielectrics of dielectric constant 4 and 6. What is the new capacitance?

    physics-General
    Initially, the capacitance of capacitor

    C equals fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction
    therefore blank fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction equals 1 mu F …(i)
    When it is filled with dielectric of dielectric constant K subscript 1 end subscript a n d blank K subscript 2 end subscript as shown, then there are two capacitors connected is parallel. So,
    C to the power of ´ ´ end exponent equals fraction numerator K subscript 1 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d end fraction plus fraction numerator K subscript 2 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d end fraction
    (as area becomes half)
    C ´ ´ blank equals fraction numerator 4 epsilon subscript 0 end subscript A over denominator 2 d end fraction plus fraction numerator 6 epsilon subscript 0 end subscript A over denominator 2 d end fraction equals fraction numerator 2 epsilon subscript 0 end subscript A over denominator d end fraction plus fraction numerator 3 epsilon subscript 0 end subscript A over denominator d end fraction
    Using Eq. (i), we obtain
    C to the power of ´ ´ end exponent equals blank 2 cross times 1 plus 3 cross times 1 equals 5 blank mu F
    General
    physics-

    The equivalent capacitance of the combination shown in figure below is

    The three capacitors are in parallel hence, their equivalent capacitance equals 3 C

    The equivalent capacitance of the combination shown in figure below is

    physics-General
    The three capacitors are in parallel hence, their equivalent capacitance equals 3 C
    General
    physics-

    The equivalent capacitance between the points A a n d B in the following circuit is

    The two capacitors each of value 1.5mu F are in parallel. So, their equivalent capacitance

    Now, three capacitors each of value 3 mu F are in series. Hence, their equivalent capacitance is given by
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 end fraction
    or fraction numerator 1 over denominator C end fraction blank equals fraction numerator 3 over denominator 3 end fraction
    or c equals 1 mu F

    The equivalent capacitance between the points A a n d B in the following circuit is

    physics-General
    The two capacitors each of value 1.5mu F are in parallel. So, their equivalent capacitance

    Now, three capacitors each of value 3 mu F are in series. Hence, their equivalent capacitance is given by
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 end fraction
    or fraction numerator 1 over denominator C end fraction blank equals fraction numerator 3 over denominator 3 end fraction
    or c equals 1 mu F
    General
    physics-

    A network of six identical capacitors, each of valueC, is made as shown in the figure.
    The equivalent capacitance between the points A space a n d blank B is

    In the given circuit capacitor’s (1) (2) and (3) are connected in series, hence equivalent capacitance is

    fraction numerator 1 over denominator C to the power of ´ ´ end exponent end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction equals fraction numerator 3 over denominator C end fraction
    rightwards double arrow C to the power of ´ ´ end exponent equals fraction numerator C over denominator 3 end fraction
    This is connected in parallel with (4).
    therefore blank C to the power of ´ ´ ´ ´ end exponent equals C to the power of ´ ´ end exponent plus C equals fraction numerator C over denominator 3 end fraction plus C equals fraction numerator 4 C over denominator 3 end fraction
    The three capacitor’s (5),fraction numerator 4 C over denominator 3 end fraction comma (6) are now connected in series.
    therefore Equivalent capacitance is
    fraction numerator 1 over denominator C blank ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 3 over denominator 4 C end fraction plus fraction numerator 1 over denominator C end fraction
    fraction numerator 1 over denominator C ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 11 over denominator 4 C end fraction
    rightwards double arrow blank C to the power of ´ ´ ´ ´ ´ ´ end exponent equals fraction numerator 4 C over denominator 11 end fraction

    A network of six identical capacitors, each of valueC, is made as shown in the figure.
    The equivalent capacitance between the points A space a n d blank B is

    physics-General
    In the given circuit capacitor’s (1) (2) and (3) are connected in series, hence equivalent capacitance is

    fraction numerator 1 over denominator C to the power of ´ ´ end exponent end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction equals fraction numerator 3 over denominator C end fraction
    rightwards double arrow C to the power of ´ ´ end exponent equals fraction numerator C over denominator 3 end fraction
    This is connected in parallel with (4).
    therefore blank C to the power of ´ ´ ´ ´ end exponent equals C to the power of ´ ´ end exponent plus C equals fraction numerator C over denominator 3 end fraction plus C equals fraction numerator 4 C over denominator 3 end fraction
    The three capacitor’s (5),fraction numerator 4 C over denominator 3 end fraction comma (6) are now connected in series.
    therefore Equivalent capacitance is
    fraction numerator 1 over denominator C blank ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 3 over denominator 4 C end fraction plus fraction numerator 1 over denominator C end fraction
    fraction numerator 1 over denominator C ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 11 over denominator 4 C end fraction
    rightwards double arrow blank C to the power of ´ ´ ´ ´ ´ ´ end exponent equals fraction numerator 4 C over denominator 11 end fraction
    General
    physics-

    Two parallel plates of area A are separated by two different dielectric as shown in figure. The net capacitance is

    C equals fraction numerator epsilon subscript 0 end subscript A over denominator fraction numerator d subscript 1 end subscript over denominator K subscript 1 end subscript end fraction plus fraction numerator d subscript 2 end subscript over denominator K subscript 2 end subscript end fraction end fraction
    equals blank fraction numerator epsilon subscript 0 end subscript A over denominator fraction numerator d over denominator 2 end fraction open parentheses fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 2 end fraction close parentheses end fraction equals fraction numerator 4 epsilon subscript 0 end subscript A over denominator 3 d end fraction

    Two parallel plates of area A are separated by two different dielectric as shown in figure. The net capacitance is

    physics-General
    C equals fraction numerator epsilon subscript 0 end subscript A over denominator fraction numerator d subscript 1 end subscript over denominator K subscript 1 end subscript end fraction plus fraction numerator d subscript 2 end subscript over denominator K subscript 2 end subscript end fraction end fraction
    equals blank fraction numerator epsilon subscript 0 end subscript A over denominator fraction numerator d over denominator 2 end fraction open parentheses fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 2 end fraction close parentheses end fraction equals fraction numerator 4 epsilon subscript 0 end subscript A over denominator 3 d end fraction