JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 23)
In the following circuit, the switch S is closed at t = 0. The charge on the capacitor C1 as a function of time will be given by $$\left( {{C_{eq}} = {{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$$
_16th_April_Morning_Slot_en_23_1.png)
$${C_1}E\left[ {1 - \exp \left( { - tR/{C_1}} \right)} \right]$$
$${C_2}E\left[ {1 - \exp \left( { - t/R{C_2}} \right)} \right]$$
$${C_{eq}}E\left[ {1 - \exp \left( { - t/R{C_{eq}}} \right)} \right]$$
$${C_{eq}}E\,\,\exp \left( { - t/R{C_{eq}}} \right)$$
Explanation
Charge on capacitor at time t,
q = q0 [ 1 $$-$$ e$$-$$t/RCeq]
= Ceq E [ 1 $$-$$ e$$-$$t/RCeq]
[ as $$\,\,\,\,\,\,$$ q0 = Ceq E]
this charge q will be on both capacitor C1 and C2, as both are in series.
q = q0 [ 1 $$-$$ e$$-$$t/RCeq]
= Ceq E [ 1 $$-$$ e$$-$$t/RCeq]
[ as $$\,\,\,\,\,\,$$ q0 = Ceq E]
this charge q will be on both capacitor C1 and C2, as both are in series.
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