Two capacitors of capacitances C1 and C2 are connected in series, assume that C1

Question

Two capacitors of capacitances C1 and C2 are connected in series, assume that C1< C2. The equivalent capacitance of this arrangement is C, where (A) C< C1 / 2 (B) C1 / 2< C < C2 / 2 (C) C1< C< C2 (D) C2< C< 2 C2

Tardigrade Admin · Accepted Answer

C1< C2 ∴ (C1/C1+C2)< (1/2) text and (C2/C1+C2) >(1/2) C=(C1 C2/C1+C2)=C1 ⋅ (C2/C1+C2) >(C1/2) Similarly, C< (C2/2)

Q. Two capacitors of capacitances $C_{1}$ and $C_{2}$ are connected in series, assume that $C_{1} < C_{2}$ . The equivalent capacitance of this arrangement is $C$ , where

$C < C_{1} /2$

$C_{1} /2 < C < C_{2} /2$

$C_{1} < C < C_{2}$

$C_{2} < C < 2 C_{2}$

$C_{1} < C_{2}$
$∴ \frac{C _{1}}{C _{1} + C _{2}} < \frac{1}{2} and \frac{C _{2}}{C _{1} + C _{2}} > \frac{1}{2}$
$C = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = C_{1} \cdot \frac{C _{2}}{C _{1} + C _{2}} > \frac{C _{1}}{2}$
Similarly, $C < \frac{C _{2}}{2}$

Q. Two capacitors of capacitances C1​ and C2​ are connected in series, assume that C1​<C2​. The equivalent capacitance of this arrangement is C, where

C<C1​/2

C1​/2<C<C2​/2

C1​<C<C2​

C2​<C<2C2​