Page 514 problems
1. Refer to fig. 20-23 for the circuit diagram.
Potential will drop due to the armature resistance
before the current actually leaves the generator.
Figure this drop using Ohm's law. Potential
across circuit will be what's left. (24V)
2. First review efficiency if you've forgotten
how it's calculated (p. 119) Since we are
given power losses, it will be easier to
figure efficiency in terms of power rather
than work done. Power is just work/time,
the time is the same for both the input and
the output power so efficiency is just power
supplied to load divided by total power produced
(times 100).
3.(a) Refer to fig. 20-24. (b) Generator
emf must equal voltage across load plus voltage
drop due to armature resistance. You'll have
to calculate armature current to figure its
voltage drop and to do this you'll have to
figure load current and field current. (252V)
(c) Read the paragraph right next to fig
20-24.
4. Use eqn in middle of p. 510, referring
to fig. 20-28.(b: 0.46 Nm)
Page 523 problems
1. Use mutual inductance eqn. Figure rate
of change of current, multiply times inductance.
3. Ratio of turns equals ratio of voltages.
4. (a) Use self inductance eqn; induced emf
will equal applied emf at instant it is applied.
(2.2H) (b) Use same rate of current change
with new induced emf, find mutual inductance.
6. (a) ratio of turns equals ratio of voltages
(b) Use Ohm's law (1.60 A) (c) Refer to p.
453 for refresher on how power is calculated
if necessary. (d) Since "losses are
negligible" ratio of currents will be
the inverse ratio of the turns.
8. (a) Ratio of voltages equals ratio of
turns (b) Losses are not negligible here.
Page 521 discusses transformer efficiency.
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