The equivalent capacitance of three capacitors, 8 mu F, and 24 mu F connected in
ID: 1517540 • Letter: T
Question
The equivalent capacitance of three capacitors, 8 mu F, and 24 mu F connected in series is 3.4 mu F 4.4 mu F 2.4 mu F 40 mu F An air-filled parallel-plate has a capacitance of 4 pF. The plate separation is then double and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes 12 pF. The dielectric constant of the wax is: 3.0 6.0 4.0 5.0 A 80 pF capacitor is charged to a potential of 2000 V. How much charge is accumulated on each plate of the capacitor? 1600 C 1250 C 1.6 times 10^-7 C 8.0 times 10^-6 C If the charge on a capacitor is doubled, its stored energy is halved is quartered is doubled is quadrupoledExplanation / Answer
22. B
given capacitors C1= 8*10^-6 F, C2= 16*10^-6 F, C3= 24*10^-6 F Connected in series
Equivalent capacitance C is given as 1/C= 1/C1 + 1/C2 + 1/C3
SUBSTITUTING THE DATA
1/C = 1/ 8*10^-6 + 1/16*10^-6 + 1/24*10^-6
C = 4.364 = 4.4 micro farad -------------> Ans option (b)
23. B
capacitance of a parallel plate capacitor is C = epsilon not*A/d
given C air filled capacitor capacitance C = 4*10^-12 F
if the capacitor is filled with dielectric material of k its constant ,its capacitance increases by k times
here d is increased by 2 times so C' = k (epsilon not A/2d)
C' = (k/2)(C)
k = C'/2C = 12*10^-12*2/4*10^-12
k = 6
dielectric constant of the wax is k = 6
24.C
From the relation between charge , capacitance, and potential is
Q = cV
Q = 80*10^-12*2000
Q = 1.6*10^-7 C ------------>Ans option C
25. D
energy stored in a capacitor is U = 1/2 cv^2 = Q^2/2C
U proportional to square of the charge
if charge is doubled then ,the stored energy becomes 4 times, quadrupoled --------->Ans
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