Two capacitors C-1 and C-2 are connected in parallel- Assume that C-1 - C-2- The equivalent capacitance of this arrangement is C- whereC - C-1-2C-1 - C - C-1C-1 - C - C-2C-2 - C - 2C-2

Question

Toppr Admin · Accepted Answer

We have- parallel capacitance- CP-C1-C2Given C2-gt-C1-x27F9-2C2-gt-CP-gt-C1-C2-gt-C2-gt-C1Hence option -a- -b- and -c- are wrong-

Two capacitors $C_{1}$ and $C_{2}$ are connected in parallel. Assume that $C_{1} < C_{2}$ . The equivalent capacitance of this arrangement is $C$ , where
$C < C_{1} / 2$
$C_{1} < C < C_{1}$
$C_{1} < C < C_{2}$
$C_{2} < C < 2 C_{2}$

We have, parallel capacitance, $C_{P} = C_{1} + C_{2}$
Given $C_{2} > C_{1}$
$⟹ 2 C_{2} > C_{P} > (C_{1} + C_{2}) > C_{2} > C_{1}$
Hence option (a), (b) and (c) are wrong.

Two capacitors C1 and C2 are connected in parallel. Assume that C1<C2. The equivalent capacitance of this arrangement is C, whereC<C1/2C1<C<C1C1<C<C2C2<C<2C2

We have, parallel capacitance, CP=C1+C2Given C2>C1⟹2C2>CP>(C1+C2)>C2>C1Hence option (a), (b) and (c) are wrong.

Two capacitors $C_{1}$ and $C_{2}$ are connected in parallel. Assume that $C_{1} < C_{2}$ . The equivalent capacitance of this arrangement is $C$ , where
$C < C_{1} / 2$
$C_{1} < C < C_{1}$
$C_{1} < C < C_{2}$
$C_{2} < C < 2 C_{2}$

We have, parallel capacitance, $C_{P} = C_{1} + C_{2}$
Given $C_{2} > C_{1}$
$⟹ 2 C_{2} > C_{P} > (C_{1} + C_{2}) > C_{2} > C_{1}$
Hence option (a), (b) and (c) are wrong.