By Theodore Korneff (Auth.)

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Additional info for Introduction to Electronics

Example text

Since there are three unknowns, we need three inde­ pendent equations. T h e third equation is found b y utilizing the other of KirchhofFs circuital laws. 4. £ Ii = 0 at any point in the circuit. Selecting a meaningful point such as B or E} we obtain Ii + h - It = 0 T h e equation £ Ii = 0 is an algebraic equation, and signs for the Ii must b e considered. A consistent choice, for example, would b e one in which currents entering a point were plus and currents leaving a point were minus. 5. T h e three independent equations chosen are Ii+ It - h = 0 - 2 J i + 3 7 2+ 10 = 0 -4/3 - 5/3 - 372= 0 I wv—i—vw 1 28 / D C Circuits Rearranging the equations for determinant use, h + h - li = -2/i + 3I2 - = 0 3/2 - 9 J -10 3 = 0 and using Cramer's rule for solving simultaneous equations, /1 I2 = = 0 -10 0 1 3 -3 -1 0 -9 1 -2 0 1 3 -3 -1 0 -9 1 -2 0 51 = 0 amp -1 0 -10 0 0 -9 90 - - a m p 51 1 -2 h 120 1 3 -3 0 -10 0 30 -amp 51 6.

10 The sum of two or more sinusoids, of the same frequency, is another sinusoid. Also, e = eL + ec + eR, the instantaneous applied voltage. This can b e obtained graphically b y adding eL, ec, and e , point R b y point, t o obtain the curve labeled e. T h e result is a sinusoid. It can be seen that e and i are o u t of phase b y the angle 6, given b y 6 = arctan [ ( X l — Xc)/R^ T h e voltages ec and eL are seen to o p p o s e each other, being a half-cycle out of phase with each other. 7 Phasor Concept Using trigonometric identities to aid in adding sinusoidal quanti­ ties that are o u t of phase w i t h each other, such as e = eL + ec + CR in the preceding section, is laborious.

F r o m n o w o n w e shall discontinue the graphical and trigono­ metric solutions and use the phasor diagrams and v e c t o r algebra. T h e sinusoids can always b e constructed, given the correct phasor diagram. 8 Average Power w h e n Phase A n g l e Is N o t Zero For the case where the circuit has an impedance for a load, the applied voltage and resultant current in general are not in phase. T h e question then is, what is the average power in this case? 67) / o Expanding, r "av = R 2 cos (9 / - .