PROBLEMS 821(b) The initial speed, final speed, and the speed
drop in r/min.
16.2.12Consider a separately excited dc motor having
a constant field current and a constant applied
armature voltage. It is accelerating a pure-inertia
load from rest. Neglecting the armature inductance
and the rotational losses, show that by the time the
motor reaches its final speed, the energy dissipated
in the armature resistance is equal to the energy
stored in the rotating parts.
16.2.13Determine the parameters of the analog capacitive
circuit shown in Figure E16.2.1(d) for the motor
in Problem 16.2.3 and its connected load. With the
aid of the equivalent circuit, obtain the expression
for the armature current, with a 3.5-starting
resistance included in series with the armature to
limit the starting current.
*16.2.14Neglecting the self-inductance of the armature cir-
cuit, show that the time constant of the equivalent
capacitive circuit for a separately excited dc mo-
tor with no load isRaReqCeq/(Ra+Req), where
Req= 1 /Geq, as shown in Figure E16.2.1(d).
16.2.15Consider the dc motor of Problem 16.2.3 to be
operating at rated voltage in steady state with a
field current of 1 A, and with the starting resistance
in series with the armature reduced to zero.
(a) Obtain the equivalent capacitive circuit ne-
glecting the armature self-inductance and cal-
culate the steady armature current.
(b) If the field current is suddenly reduced to 0.8 A
while the armature applied voltage is constant
at 220 V, compute the initial armature current
ia(0) on the basis that the kinetic energy stored
in the rotating parts cannot change instanta-
neously.
(c) Determine the final armature currentia(∞) for
the condition of part (b).
(d) Obtain the time constantτam′ of the armature
current for the condition of part (b), and ex-
press the armature current as a function of
time on the basis thatia=ia(∞)+[ia( 0 )−
ia(∞)]e−t/τ
am′
.
16.2.16A separately excited dc motor, having a constant
field current, accelerates a pure inertia load from
rest. If the system is represented by an electrical
equivalent circuit, with symbols as shown in Fig-
ure P16.2.16, expressR, L, andCin terms of the
motor parameters.
16.2.17Figure P16.2.17 represents the Ward–Leonard sys-
tem for controlling the speed of the motorM.
With the generator field voltagevfgas the input
and the motor speedωmas the output, obtain an
expression for the transfer function for the system,
assuming idealized machines. Let the load on the
motor be given byJω ̇m+Bωm. The generator
runs at constant angular velocityωg.
16.2.18A separately excited dc generator has the follow-
ing parameters:Rf= 100 , Ra= 0. 25 , Lf=
25 H,La= 0 .02 H, andKg=100 V per field
ampere at rated speed.
(a) The generator is driven at rated speed, and a
field circuit voltage ofVf=200 V is suddenly
applied to the field winding.
(i) Find the armature generated voltage as a
function of time.
(ii) Calculate the steady-state armature volt-
age.
(iii) How much time is required for the arma-
ture voltage to rise to 90% of its steady-
state value?
(b) The generator is driven at rated speed, and
a load consisting ofRL= 1 andLL =
0 .15 H in series is connected to the armature
terminals. A field circuit voltageVf = 200
V is suddenly applied to the field winding.
Determine the armature current as a function
of time.J
If = constant(a) (b)La Ra i
a R i Le v C
ωm+−v+−−+Figure P16.2.16