(Volt) V(t)K (Volt)...M (Volt).-.... K (Volt)(Amper)i(t)C(Amper)(saniye)01801С (Аmper)In the figure given above, the variation of the(saniye)150voltage between the terminals of a thyristor(period) is given. Calculate the powerdissipated by this thyristor during T time, inWatts , using the values given below :T= 20 ms52189b18М 3 3 Voltsb180100558 - 9952189115b18C = 20 Ampsb180100558 - 9952189115091150b18180100558 9521891150BA95.699115u68.92b180100558 - 99521891150NS206.56180100558 99521тоb18138.93183.21b180100558 - 9952189115ub18801005581801005580k

From given graph we will find expression for voltage and current through the thyristor. power…

(Volt) v(t) K(Volt).-.- M(Volt). .. K (Volt) (Amper) i„(t) C(Amper) (eaniye) 1801 CLAmper) (saniye) In the figure given above, the variation of the voltage between the terminals of a thyristor (period) is given. Calculate the power dissipated by this thyristor during T time, in Watts , using T= 20 ms b18 Volts M = 3 Volts b180100558 - 9952189115u b18 C = 20 Amps b180100558 - 9952189115u 91150 b18 A 9115 95.69 180100558 952189115 68.92 b180100558 - 9952189115 NS 150 b18 138.93 то 183.21 b180100558 - 9952189115u b18 1150

(Volt) v(t) K(Volt).-.- M(Volt). .. K (Volt) (Amper) i„(t) C(Amper) (eaniye) 1801 CLAmper) (saniye) In the figure given above, the variation of the voltage between the terminals of a thyristor (period) is given. Calculate the power dissipated by this thyristor during T time, in Watts , using T= 20 ms b18 Volts M = 3 Volts b180100558 - 9952189115u b18 C = 20 Amps b180100558 - 9952189115u 91150 b18 A 9115 95.69 180100558 952189115 68.92 b180100558 - 9952189115 NS 150 b18 138.93 то 183.21 b180100558 - 9952189115u b18 1150

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Concept explainers

Distribution System

The final stage in the delivery of the electric power is the electric power distribution system. From the transmission system to the individual customers it carries electricity. The distribution system is connected with the transmission system and is used to lower the voltage of the transmission to medium voltage in the range of 2 kV to 35 kV with the help of transformers. This medium voltage to the distribution transformers located near the customer's premises is carried by the primary distribution system lines. These distribution transformers in turn lower the voltage to utilization voltage used in lightings, household appliances, industrial equipments etc. Through secondary distribution lines often several customers are supplied from one transformer. It is through service drops, the commercial and residential customers are connected to the secondary distribution system lines. Customers who demand a larger amount of power can be connected directly to the primary distribution system level or the sub-transmission level.

Question

(Volt) V(t)
K (Volt)...
M (Volt).
-.... K (Volt)
(Amper)
i(t)
C(Amper)
(saniye)
01801
С (Аmper)
In the figure given above, the variation of the
(saniye)
150
voltage between the terminals of a thyristor
(period) is given. Calculate the power
dissipated by this thyristor during T time, in
Watts , using the values given below :
T= 20 ms
52189
b18
М 3 3 Volts
b180100558 - 9952189115
b18
C = 20 Amps
b180100558 - 99521891150
91150
b18
180100558 9521891150
B
A
95.69
9115u
68.92
b180100558 - 99521891150
NS
206.56
180100558 99521
то
b18
138.93
183.21
b180100558 - 9952189115u
b18
80100558
1801005580k

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