1. Singly and doubly excited magnetic systems applications are respectively ________
a) loud speakers and tachometers
b) synchronous motors and moving iron instruments
c) DC shunt machines and solenoids
d) reluctance motors and synchronous motors
Explanation: Reluctance motors can work on single excitation, and synchronous motors require double excitation.
2. Electromagnetic torque in rotating electrical machinery is present when ________
a) stator winding alone carries current
b) rotor winding alone carries current
c) air gap is uniform
d) both stator and rotor windings carry current
Explanation: Electromagnetic torque = isirdMsr/dθr and if either is or ir = 0, then Te=0.
3. All practical electromechanical energy conversion devices make use of the magnetic field rather than the electric field as the coupling medium. This is because ________
a) electric field systems present insulation difficulties
b) electric field systems have more dielectric loss than the magnetic loss, for the same power rating of the machine
c) in electric field systems, for normal electric field strength, the stored energy density is high
d) in magnetic field systems, for normal magnetic flux density, the stored energy density is high
Explanation: As the energy storing capacity of the magnetic field is higher, magnetic field as coupling medium is most common in electromechanical energy conversion devices.
4. 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
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
5. 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)
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
6. 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)
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.
7. 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
Explanation: A2=2A1, x2=2x1, v2=v1, fe2=1/2v2dC(x)/dx2, C2=ε0A2/x2 ⇒ dC2/dx2 = -ε0A2/x22
dC2/dx2 = -2ε0A1/4x12 = -ε0A1/2x12 ⇒ f&e2 = -1/2v12ε0A1/2x12 = -1/2(fe1)
⇒ force reduces to half.
8. 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
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
9.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)
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 D2/ε0 and as D is increasing, Wfld increases.
10. 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
Explanation: fe = -∂Wfld(q,x)/∂x
fe = -∂Wfld1(q,x)/∂x.