By C.L. Wadhwa
Easy electric Engineering Has Been Written As A center direction For All Engineering scholars Viz. Electronics And verbal exchange Engineering, desktop Engineering, Civil Engineering, Mechanical Engineering and so forth. because This path Will in general Be provided on the First yr point Of Engineering, the writer Has Made Modest attempt to offer In A Concise shape, a variety of beneficial properties Of uncomplicated electric Engineering utilizing uncomplicated Language and during Solved Examples, warding off The Rigorous Of Mathematics.The Salient beneficial properties Of The publication Are:
* regular kingdom research Of A.C. Circuits defined.
* community Theorems defined utilizing general Examples.
* research Of 3-Phase Circuits And size Of strength In those Circuits defined.
* Measuring tools Like Ammeter, Voltmeter, Wattmeter and effort Meter defined.
* quite a few electric Machines Viz. Transformers, D.C. Machines, unmarried section and 3 part Induction vehicles, Synchronous Machines, Servomotors were defined.
* a quick View Of energy procedure together with traditional And Non-Conventional prone of electrical power Is Given.
* family Wiring Has Been mentioned.
* a number of Solved Examples And perform difficulties For Thorough grab Of the topic awarded.
* a lot of a number of selection Questions With solutions Given.
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Additional info for Basic Electrical Engineering
Id=10367724&ppg=62 - Solution.
A resistance is connected in series with an unknown resistance R and R 20 n �-----41-----l 1 1 t----� 220 V Fig. 11 Solution. 5 Q Let us find out which one of these a valid result. 5 = 57 watts. 5 Q is more appropriate. Series resistance as mentioned earlier is used to limit the current in the circuit. Suppose we have an appliance rated for 1 10 volts . However, the source voltage available is 220 volts. The device can't be connected directly to the supply. Hence a resistance in series with the device is connected and then the combination put across the supply and by adjusting the series resistance 1 1 0 volts can be adjusted across the device.
Ans. Hence Voltage drop across and power dissipated in each resistor can be obtained. Example e 4 [l i. 1 7. - - Determine the current in each branch of the network shown. d 4 [l c an 150 4 0 v -=- a 3 0 (a) (d) 150 40 V -=- 10n 20 60 b (b) (e) (c) Fig. 1 7 Solution. The various steps have been shown in the figures above. Fig. Ei. 17 (b) is ob tained by finding equivalent of parallel resistance 3 n and 6 n 3x6 ,.... = 2 �� i. e. 3+ 6 In Fig. i. 1 7 (c) two series resistances 2 n and 8 n have been combined.