III. Multiple capacitors A bulb is connected to a battery and two capacitors as
ID: 1594852 • Letter: I
Question
III. Multiple capacitors A bulb is connected to a battery and two capacitors as shown at right Suppose that C, is less than C, A. Before connecting the circuit a student makes the following prediction: "Current flows from the positive side of the battery to the negative side of the battery. Since the bulb is isolated from the battery on both sides by the capacitors, the bulb will not light." Do you agree or disagree with this prediction? Explain. Obtain a second capacitor from your instructors and check your prediction.Explanation / Answer
(A) Bulb will light becuse the circuit is complete and as given that current flows from positive to negative end , currrent can only flow when circuit is complete.
(B)
(1 ) Just after switch is closed so both the capacitor is not charged hence there is no potential across the cpacitors. whole current is flowing through the bulb as if there is only bulb in the circuit .
brightness of bulb will be maximum.
(.) potential differnece across capacitor is zero
as there is no charge Q=0
so V=Q/c
V=0
(2)
(.) When the switch is closed for long time both the capacitor is fully charged so now there is potential differnce accros the capacitor potential difference is
as potential is inversly proportional to capacitance
V1= (C2V)/(C1 +C2) (where V is potential of battery)
and
V2=(C1V)/(C1 +C2)
so Potential diffence across bulb is
V2-V1= V(C1-C2)/(C1 +C2) ANS
(.)
V1+V2=V(C2+C1)/(C1 +C2)
V1+V2=V
(.)
Yes final charge is same on C1 and C2 , as in seris combination each capacitor has same charge .
Q1 =C1 V1
Q1 =C1(C2V)/(C1+C2)
and Q2=C2V2
Q2=C2(C1V)/(C1+C2)
so
Q1=Q2
(.)
potential difference on C1 and C2 will depend on the capacitance of the capacitor. as potential is inversly proportional to capacitance
if C1>C2
then V1<V2
If C1<C2
then V1>V2 ANS