Electrical Machines Questions and Answers_pag16

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Explanation: C= εA/x and as separation is decreased, capacitance increases. Also, C=q/v and 
as charge remains constant, voltage decreases. Similarly, energy stored, W​fld​=1/2q​2​/c(x) and it 
decreases. 
3. A parallel plate capacitor is charged and then the DC supply is disconnected. The plate 
separation is then increased. Between the plates, 
(i) electric field intensity is unchanged 
(ii) flux density decreases 
(iii) potential difference decreases 
(iv) energy stored increases 
Which of the above statements are correct? 
a) (i),(iv) 
b) (ii),(iv) 
c) (ii),(iii),(iv) 
d) (i),(iii),(iv) 
View Answer 
Answer: a 
Explanation: We know C=ε​0​A/x, as x increases, C decreases. 
Charge(q) on plate remains constant and q=Cv implies v increases. 
Energy stored, W​fld​= 1/2q​2​/C(x), as C decreases, W​fld​ increases. 
Similarly as q=DA and q,A remains constant, D doesn’t change and as electric field intensity is 
given by, E=D/ε​0​, E also doesn’t change. 
4. The area of two parallel plates is doubled and the distance between these plates is also 
doubled. The capacitor voltage is kept constant. Under these conditions, force between the 
plates of this capacitor __________ 
a) decreases 
b) increases 
c) reduce to half 
d) gets doubled 
View Answer 
Answer: c 
Explanation: A​2​=2A​1​, x​2​=2x​1​, v​2​=v​1​, f​e2​=1/2v​2​dC(x)/dx​2​, C​2​=ε​0​A​2​/x​2​ ⇒ dC​2​/dx​2​= -ε​0​A​2​/x​2​
2 
dC​2​/dx​2​= -2ε​0​A​1​/4x​1​
2​ = -ε​0​A​1​/2x​1​
2​ ⇒ f&​e2​= -1/2v​1​
2​ε​0​A​1​/2x​1​
2​= -1/2(f​e1​) 
⇒ force reduces to half. 
5. A parallel plate capacitor has an electrode area of 1000 mm​2​, with a spacing of 0.1 mm 
between the electrodes. The dielectric between the plates is air with a permittivity of 8.85∗10​-12 
F/m. The charge on the capacitor is 100 v. The stored energy in the capacitor is ____________ 
a) 44.3 J 
b) 444.3 nJ 
c) 88.6 nJ 
d) 44.3 nJ 
View Answer 
Answer: d 
Explanation: W​fld​ = 1/2 q​2​/C(x)