General
Easy
Physics-

A particle starts from rest at t equals 0 and undergoes an acceleration a in m s to the power of negative 2 end exponent with time t in seconds which is as shown

Which one of the following plot represents velocity V in m s to the power of negative 1 end exponent versus time t in seconds

Physics-General

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    Answer:The correct answer is:
    Takingthe motion from 0 to 2 blank s
    u equals 0 comma blank a equals 3 m s to the power of negative 2 end exponent comma blank t equals 2 s comma blank v equals ?
    v equals u plus a t equals 0 plus 3 cross times 2 equals 6 m s to the power of negative 1 end exponent
    Taking the motion from 2 blank s to 4 blank s
    v equals 6 plus open parentheses negative 3 close parentheses open parentheses 2 close parentheses equals 0 m s to the power of negative 1 end exponent

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    Period of cos to the power of 6 space x plus sin to the power of 6 space x is

    Period of cos to the power of 6 space x plus sin to the power of 6 space x is

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    Two inclined planes are located as shown in figure. A particle is projected from the foot of one frictionless plane along its line with a velocity just sufficient to carry it to top after which the particle slides down the other frictionless inclined plane. The total time it will take to reach the point C is

    The time of ascent = time of descent equals t subscript 0 end subscript
    T equals total time of flight equals 2 t subscript 0 end subscript

    sin invisible function application 45 degree equals fraction numerator 9.8 over denominator B C end fraction equals fraction numerator 9.8 over denominator s end fraction
    therefore blank s equals 9.8 square root of 2
    therefore blank s equals u t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent
    s equals 0 cross times t plus fraction numerator 1 over denominator 2 end fraction left parenthesis g sin invisible function application 45 degree right parenthesis t subscript 0 end subscript superscript 2 end superscript
    or 9.8 square root of 2 equals fraction numerator 9.8 over denominator 2 square root of 2 end fraction t subscript 0 end subscript superscript 2 end superscript
    therefore blank t subscript 0 end subscript superscript 2 end superscript equals 4
    therefore blank t subscript 0 end subscript equals 2 s
    therefore blank T equals 2 t subscript 0 end subscript minus 4 s

    Two inclined planes are located as shown in figure. A particle is projected from the foot of one frictionless plane along its line with a velocity just sufficient to carry it to top after which the particle slides down the other frictionless inclined plane. The total time it will take to reach the point C is

    physics-General
    The time of ascent = time of descent equals t subscript 0 end subscript
    T equals total time of flight equals 2 t subscript 0 end subscript

    sin invisible function application 45 degree equals fraction numerator 9.8 over denominator B C end fraction equals fraction numerator 9.8 over denominator s end fraction
    therefore blank s equals 9.8 square root of 2
    therefore blank s equals u t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent
    s equals 0 cross times t plus fraction numerator 1 over denominator 2 end fraction left parenthesis g sin invisible function application 45 degree right parenthesis t subscript 0 end subscript superscript 2 end superscript
    or 9.8 square root of 2 equals fraction numerator 9.8 over denominator 2 square root of 2 end fraction t subscript 0 end subscript superscript 2 end superscript
    therefore blank t subscript 0 end subscript superscript 2 end superscript equals 4
    therefore blank t subscript 0 end subscript equals 2 s
    therefore blank T equals 2 t subscript 0 end subscript minus 4 s
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    I minus Vcharacteristic of a copper wire of length L and area of cross-section A is shown in figure. The slope of the curve becomes

    Slope of graph
    equals fraction numerator I over denominator V end fraction equals fraction numerator 1 over denominator R end fraction
    If experiment is performed at higher temperature then resistance increase and hence slope decrease, choice (a) is wrong.
    Similarly in choice (b) and (c) resistance increase.
    But for choice (d) resistance R increases, so slope decreases

    I minus Vcharacteristic of a copper wire of length L and area of cross-section A is shown in figure. The slope of the curve becomes

    physics-General
    Slope of graph
    equals fraction numerator I over denominator V end fraction equals fraction numerator 1 over denominator R end fraction
    If experiment is performed at higher temperature then resistance increase and hence slope decrease, choice (a) is wrong.
    Similarly in choice (b) and (c) resistance increase.
    But for choice (d) resistance R increases, so slope decreases
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    Four concurrent coplanar forces in newton are acting at a point and keep it in equilibrium figure. Then values of F and theta are

    In equilibrium position along y-direction
    2 sin 60degree equals square root of 3 plus F cos invisible function application theta
    or 2 cross times fraction numerator square root of 3 over denominator 2 end fraction equals square root of 3 plus F cos invisible function application theta or F cos invisible function application theta equals 0
    As F not equal to 0
    therefore blank cos invisible function application theta equals 0 or theta equals 90 degree
    Along x-direction, F blank s i n 90 degree equals 1 plus 2 c o s 60 degree
    equals 1 plus 2 cross times fraction numerator 1 over denominator 2 end fraction
    F equals 2N

    Four concurrent coplanar forces in newton are acting at a point and keep it in equilibrium figure. Then values of F and theta are

    physics-General
    In equilibrium position along y-direction
    2 sin 60degree equals square root of 3 plus F cos invisible function application theta
    or 2 cross times fraction numerator square root of 3 over denominator 2 end fraction equals square root of 3 plus F cos invisible function application theta or F cos invisible function application theta equals 0
    As F not equal to 0
    therefore blank cos invisible function application theta equals 0 or theta equals 90 degree
    Along x-direction, F blank s i n 90 degree equals 1 plus 2 c o s 60 degree
    equals 1 plus 2 cross times fraction numerator 1 over denominator 2 end fraction
    F equals 2N
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    The effective capacitance between points X and Y shown in figure. Assuming C subscript 2 end subscript equals 10 blank muF and that outer capacitors are all 4 blank muF is

    The arrangement shows a Wheatstone bridge.
    As fraction numerator C subscript 1 end subscript over denominator C subscript 3 end subscript end fraction equals fraction numerator C subscript 4 end subscript over denominator C subscript 5 end subscript end fraction equals 1 comma blanktherefore the bridge is balanced.
    fraction numerator 1 over denominator C subscript s subscript 1 end subscript end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 2 over denominator 4 end fraction equals fraction numerator 1 over denominator 2 end fraction comma C subscript s subscript 1 end subscript end subscript equals 2 mu blank F
    Similarly, C subscript s end subscript subscript 2 end subscript equals 2 mu blank F
    therefore effective capacitance
    equals C subscript p end subscript equals C subscript s end subscript subscript 1 end subscript plus C subscript s end subscript subscript 2 end subscript equals 2 plus 2 plus equals 4 mu blank F

    The effective capacitance between points X and Y shown in figure. Assuming C subscript 2 end subscript equals 10 blank muF and that outer capacitors are all 4 blank muF is

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    The arrangement shows a Wheatstone bridge.
    As fraction numerator C subscript 1 end subscript over denominator C subscript 3 end subscript end fraction equals fraction numerator C subscript 4 end subscript over denominator C subscript 5 end subscript end fraction equals 1 comma blanktherefore the bridge is balanced.
    fraction numerator 1 over denominator C subscript s subscript 1 end subscript end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 2 over denominator 4 end fraction equals fraction numerator 1 over denominator 2 end fraction comma C subscript s subscript 1 end subscript end subscript equals 2 mu blank F
    Similarly, C subscript s end subscript subscript 2 end subscript equals 2 mu blank F
    therefore effective capacitance
    equals C subscript p end subscript equals C subscript s end subscript subscript 1 end subscript plus C subscript s end subscript subscript 2 end subscript equals 2 plus 2 plus equals 4 mu blank F
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    physics-

    The four capacitors, each of 25 muF are connected as shown in figure. The DC voltmeter reads 200 V. the change on each plate of capacitor is

    Charge on each plate of each capacitor
    Q equals plus-or-minus C V equals plus-or-minus 25 cross times 10 to the power of negative 6 end exponent cross times 200
    equals plus-or-minus 5 cross times 10 to the power of negative 3 end exponent C

    The four capacitors, each of 25 muF are connected as shown in figure. The DC voltmeter reads 200 V. the change on each plate of capacitor is

    physics-General
    Charge on each plate of each capacitor
    Q equals plus-or-minus C V equals plus-or-minus 25 cross times 10 to the power of negative 6 end exponent cross times 200
    equals plus-or-minus 5 cross times 10 to the power of negative 3 end exponent C
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    For the circuit shown in figure the charge on 4muF capacitor is

    Combined capacity of 1 blank mu F and 5 mu F blank= 1 + 5=6 blank mu F
    Now, 4 mu F blankand 6 mu F are in series.
    therefore blank fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 6 end fraction plus fraction numerator 3 plus 2 over denominator 12 end fraction equals fraction numerator 5 over denominator 12 end fraction
    C subscript s end subscript equals fraction numerator 12 over denominator 5 end fraction mu F
    Charge in the arm containing 4 mu F blankcapacitor is
    q equals C subscript s end subscript cross times V equals fraction numerator 12 over denominator 5 end fraction cross times 10 equals 24 blank mu C

    For the circuit shown in figure the charge on 4muF capacitor is

    physics-General
    Combined capacity of 1 blank mu F and 5 mu F blank= 1 + 5=6 blank mu F
    Now, 4 mu F blankand 6 mu F are in series.
    therefore blank fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 6 end fraction plus fraction numerator 3 plus 2 over denominator 12 end fraction equals fraction numerator 5 over denominator 12 end fraction
    C subscript s end subscript equals fraction numerator 12 over denominator 5 end fraction mu F
    Charge in the arm containing 4 mu F blankcapacitor is
    q equals C subscript s end subscript cross times V equals fraction numerator 12 over denominator 5 end fraction cross times 10 equals 24 blank mu C
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    The variation of electric potential with distance from a fixed point is shown in figure. What is the value of electric field at x=2 m.

    The variation of electric potential with distance from a fixed point is shown in figure. What is the value of electric field at x=2 m.

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    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

    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

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    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