Question 359: Electrostatics & Capacitance

Question

In the following circuit,  the switch S is closed at t = 0. The charge on the capacitor C
1
as a function of time will be given by $\left( {{C_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l8g3fvur/5cac09d0-17a2-49f4-9e34-1b454eb23291/5882fc40-3c21-11ed-8acd-3bfc6080b9d7/file-1l8g3fvus.png?format=png)

Accepted Answer

Charge on capacitor at time t,
q = q
0
[ 1 $-$ e
$-$t/RC
eq
]
= C
eq
E [ 1 $-$ e
$-$t/RC
eq
]
[ as $\,\,\,\,\,\,$ q
0
= C
eq
E]
this charge q will be on both capacitor C
1
and C
2
, as both are in series.

Answer

${C_1}E\left[ {1 - \exp \left( { - tR/{C_1}} \right)} \right]$

Answer

${C_2}E\left[ {1 - \exp \left( { - t/R{C_2}} \right)} \right]$

Answer

${C_{eq}}E\left[ {1 - \exp \left( { - t/R{C_{eq}}} \right)} \right]$

Answer

${C_{eq}}E\,\,\exp \left( { - t/R{C_{eq}}} \right)$

Electrostatics & Capacitance

In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C 1 as a function of time will be given by \left( {{C_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)

Choose the correct answer: