(*) Let, equivalent capacitance of two capacitors
C1 and
C2 connected in parallel be
Ca and equivalent capacitance of same, when connected in series be
Cb.
According to given data,
Ca=‌Cb . . . (i)
Since, equivalent capacitance in parallel combination,
Ceq=C1+C2 Ca=C1+C2 . . . (ii)
and equivalent capacitance in series combination,
‌‌‌=‌+‌ ‌‌‌=‌+‌=‌ Cb‌‌=‌. . . (iii)
Substituting Eqs. (ii) and (iii) in Eq. (i), we get
C1+C2=‌‌+C2 4(C1+C2)2=15C1C2 ⇒‌‌4C12+4C22+8C1C2=15C1C2 ⇒‌‌4C12+4C22−7C1C2=0 On dividing both sides by
C12, we get
4+4(‌)2−7‌=0 ‌ or ‌‌‌4(‌)2−7(‌)+4=0 ‌ If ‌‌=x,‌ then ‌4x2−7x+4=0 By using the concept of quadratic equation,
‌x=‌⇒x=‌ ⇒‌‌x=‌=‌=‌