JEE Main 26 Feb 2021 Shift 1 Solved Paper

(*) 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=‌

15

4

Cb . . . (i)
Since, equivalent capacitance in parallel combination,
Ceq=C1+C2
Ca=C1+C2 . . . (ii)
and equivalent capacitance in series combination,
‌

1

Ceq′

‌‌=‌

1

C1

+‌

1

C2

‌

1

Cb

‌‌=‌

1

C1

+‌

1

C2

=‌

C2+C1

C1C2

Cb‌‌=‌

C1C2

C1+C2

. . . (iii)
Substituting Eqs. (ii) and (iii) in Eq. (i), we get
C1+C2=‌

15

4

‌

C1C2

C1

+C2
4(C1+C2)2=15C1C2
⇒‌‌4C12+4C22+8C1C2=15C1C2
⇒‌‌4C12+4C22−7C1C2=0
On dividing both sides by C12, we get
4+4(‌

C2

C1

)2−7‌

C2

C1

=0
‌ or ‌‌‌4(‌

C2

C1

)2−7(‌

C2

C1

)+4=0
‌ If ‌‌

C2

C1

=x,‌ then ‌4x2−7x+4=0
By using the concept of quadratic equation,
‌x=‌

−b±√b2−4ac

2a

⇒x=‌

7±√49−64

8

⇒‌‌x=‌

C2

C1

=‌

7±√−15

8

=‌

7±√15i

8