Question III.B.3: (a) Derive an expression for the EMF equation of a dc genera...
(a) Derive an expression for the EMF equation of a dc generator.
(b) The applied voltage and current fl owing through in an L–R circuit are
v =200 sin (314t+π/3) and
I= 20 sin(314t+π/6)
Calculate the (i) power factor, (ii) average power, (iii) impedance, and (iv) values of R and L
(a) EMF equation of a dc generator: Figure 45 shows a number of conductors rotating in a mag-netic field. EMF will be induced in these conductors. We will deduce a generalized equation for the EMF induced in the dc generator.
Let us assume the following:
P = No. of poles of the dc generator
Z = Total number of conductors placed in the armature
Φ = Flux per pole
N = Revolution per minute of the armature
A = No. of parallel paths of the armature winding
EMF induced in a conductor = Flux cut/Second
In one revolution a conductor will cut PΦ flux. Time taken for the conductor to make one revolu-tion in\frac{60}{N} seconds
EMF induced per conductor = \frac{Flux cut}{Time taken}=\frac{P\phi}{\frac{60}{N}}=\frac{P\phi N}{60} Wbs
Z number of conductors are connected in series in A number of parallel paths. Therefore, the number of conductors in each parallel path will be \frac{Z}{A}
The induced EMF, E, in the generator is equal to the induced EMF in one parallel path.
Thus,
(b) v =200 sin (314 t+π/3)
i= 20 sin(314 t+π/6)
These quantities are shown in Fig. 46.
(i) Power factor, cos Φ = cos30° = 0.866 lagging
(ii) Average power = V I cos Φ=\frac{V_{m}}{\sqrt{2}}\frac{I_{m}}{\sqrt{2}} cos \phi
V_{m} =200 =\frac{200}{\sqrt{2}}×\frac{20}{\sqrt{2}}× 0.866
I_{m}= 20 =2000 ×0.866
=1732 W
(iii) Impedance, Z =\frac{V}{I}=\frac{\frac{V_{m}}{\sqrt{2}}|\underset{}{\underline{60°}}} {\frac{I_{m}}{\sqrt{2}}|\underset{}{\underline{30}}}=\frac{200}{20}|\underset{}{\underline{30}}
=10 |\underset{}{\underline{30º}}=10(cos 30º+ j sin 30º)
=8.66 + j5
= R + j X_{L}
(iv) R=8.66 Ω
X_{L}= \omega L =2πfL =5L =\frac{5}{2πf} = \frac{5}{314}H =0.0159=5.9 mH
