A parallel plate capacitor is connected with a 1.694 volt batten- and each plate
ID: 1512996 • Letter: A
Question
A parallel plate capacitor is connected with a 1.694 volt batten- and each plate contains 4.118 micro Coulomb charge. How much energy is stored in the capacitor? Your Answer: A certain capacitor stores 37 J of energy when it holds 2,270 uC of charge. What is the capacitance in nF? Your Answer: C_1-4F, C_2-4K, C_3-2F, C_4-4F, C_5- 11.2 F. Calculate the equivalent capacitance between A and B points. Your Answer: Two circular metal plates, separated by a distance 8.85 mm forms 55 nF capacitance. The space between the plates are filled with Tefflon material with epsilon_r = 2.2. Calculate the area of each plate. Your AnswerExplanation / Answer
Parallel plate capacitors
Given parallel plate capacitor battery is V = 1694 V, each plate have charge of Q = 4118 C
Q-9
we know that energy stored in the capacitor is U = 1/2 cV^2
and relation between charge, potential, capacitance is Q =CV => c = Q/V
U = 1/2(Q/v)V^2 = 1/2 QV
U = 0.5*4118*1694 J
U = 3487946 J
energy stored is 3487946 J
Q-10
energy stored in the capacitor is U = 1/2 Q^2/C ==> C = 0.5*Q^2/U = 0.5*(2270*10^-6)^2/37
capacitance C = 6.9634*10^-8 F
C = 69.634 nF
Q-11
given C1=4F,C2=4F,C3=2F,C4=4F, C5=11.2 F
Equivalent capacitance between A&B is
C1,C2 ; C3,C4 ARE parallel so C1,2 = C1+C2 = 4+4 = 8 F, C3,4 = C3+C4 = 2+4 = 6 F
now C12,C34,C5 all are in series so net capacitance is 1/C = 1/C12 + 1/C34 +1/C5
SUBSTITTUTING THE VALUES 1/C = 1/8 + 1/6 + 1/11.2
C = 2.625 F
so equivalent capacitance between A and Bis 2.625 F
Q12
we know that the capacitance of a capacitor is C = epsilon not *A/d
==> A = kC*d/ epsoilon not
= 2.2*55*10^-9 *8.85*10^-3 / 8.85*10^-12
= 55*2.2 = 121 m2
area of plate is 121 m2
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