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 $$\left( {{C_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$$

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: