Electrical Machines Questions and Answers – Singly Excited Electric Field Systems

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This set of Electrical Machines Multiple Choice Questions & Answers (MCQs) focuses on “Singly Excited Electric Field Systems”.

1. A parallel plate capacitor has a capacitance of 10μF. If the linear dimensions of the plates are doubled and distance between them is also doubled, the new value of capacitance would be __________
a) 10μF
b) 20μF
c) 5μF
d) 40μF
View Answer

Answer: b
Explanation: C1 = 10μ F = ε0A1/x1
C2 = ε0A2/x2, A2=4A1(as the linear dimensions are doubled, area increases by 4 times) and x2=2x1
⇒ C2 = ε04A1/2x1 = 2∗10μF = 20μF
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2. A parallel plate capacitor is changed and then the DC supply is disconnected. Now plate separation is allowed to decrease due to force of attraction between the two plates. As a consequence which of the following statements are correct?

(i) charge on the plate increases
(ii) charge on the plates remain constant 
(iii) capacitance C increases
(iv) capacitance C remains constant
(v) potential difference increases
(vi) potential difference decreases
(vii) energy stored decreases
(viii) energy stored increases

a) (i), (iii), (vi)
b) (ii), (iv), (viii)
c) (ii), (iii), (vi), (vii)
d) (ii), (iv), (v), (viii)
View Answer

Answer: c
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, Wfld=1/2q2/c(x) and it decreases.
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3. A parallel plate capacitor is charged and then the DC supply is disconnected. The plate separation is then increased. Between the plates which of the following statements are correct?

(i) electric field intensity is unchanged
(ii) flux density decreases
(iii) potential difference decreases
(iv) energy stored increases

a) (i), (iv)
b) (ii), (iv)
c) (ii), (iii), (iv)
d) (i), (iii), (iv)
View Answer

Answer: a
Explanation: We know C=ε0A/x, as x increases, C decreases.
Charge(q) on plate remains constant and q=Cv implies v increases.
Energy stored, Wfld = 1/2q2/C(x), as C decreases, Wfld increases.
Similarly as q=DA and q, A remains constant, D doesn’t change and as electric field intensity is given by, E=D/ε0, E also doesn’t change.

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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: A2=2A1, x2=2x1, v2=v1, fe2=1/2v2dC(x)/dx2, C20A2/x2 ⇒ dC2/dx2 = -ε0A2/x22
dC2/dx2 = -2ε0A1/4x12 = -ε0A1/2x12 ⇒ f&e2 = -1/2v12ε0A1/2x12 = -1/2(fe1)
⇒ force reduces to half.

5. A parallel plate capacitor has an electrode area of 1000 mm2, 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: Wfld = 1/2 q2/C(x)
C(x) = ε0A/x = 8.85∗10-12∗100∗(10-3)2/0.1∗10-3 = 8.85∗10-12 F
Wfld = 1/2∗Cv = 1/2∗1002∗8.85∗10-12 = 44.3nJ
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6. A parallel plate capacitor is connected to a DC source. Now the plates are allowed to move a small displacement under the influence of force of attraction between the two plates. As a result ____________

(i) charge on the plates increases
(ii) charge on plates remains constant
(iii) energy stored increases
(iv) energy stored remains constant
(v) electric field intensity is unchanged
(vi) flux density increases

Which of the above statements are correct?
a) (ii), (iv), (v)
b) (ii), (iii), (vi)
c) (i), (iii), (v)
d) (i), (iii), (vi)
View Answer

Answer: d
Explanation:C = ε0A/x ⇒ C ∝ 1/x ⇒ If the force if of attraction type, then x decreases and C increases.
q=Cv, v constant⇒ as C increases, q also increases. We also know that q=DA and if q increaes, D increases.
Finally, Wfld = 1/2 D20 and as D is increasing, Wfld increases.
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7. The force produced by electric field in a singly excited energy conversion device, using electric field as coupling medium can be obtained by ___________
a) use of field energy function only
b) use of coenergy function only
c) use of field energy or coenergy function
d) none of the mentioned
View Answer

Answer: c
Explanation: fe = -∂Wfld(q,x)/∂x
fe = -∂Wfld1(q,x)/∂x.

8. Charge and voltage associated with electric field are analogous, respectively to __________ and ___________ in magnetic field.
a) flux linkages and current
b) flux density and current
c) flux linkages and voltage
d) MMF and current
View Answer

Answer: a
Explanation: In magnetic field, the basic terms are flux linkages and current, which are responsible for energy conversion and in electric field, charge and voltage are basically responsible for energy conversion.
dWelec = ψi, for magnetic field
dWelec = vidt=vdq, for electric field.

9. Two parallel plates, each of area A = 1m2 are separated by a distance g. The electric field intensity between the plates is 3∗106 v/m. What is the force between the two plates?
a) 1/2π∗103 N
b) 1/8π∗103 N
c) 8π∗103 N
d) 0 N
View Answer

Answer: b
Explanation: When electric field is applied, the plates move towards each other because of force of attraction by a distance x ⇒ C = ε0A/(g-x)
Wfld(q,x) = 1/2 q2/C = 1/2 q2(g-x)/Aε0
fe = -∂Wfld(q,x)/∂x = 1/2q2x/Aε0
We know, q = DA = ε0EA and ε0 = 10-9/36Π
fe = 1/2 E2ε0A = 1/2∗(3∗106)2∗10-9/36Π∗1 = 1/8π∗103 N

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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