By R. Yorke

This revised and enlarged version offers a concise and finished creation to simple recommendations ordinary within the fields of electric, digital and keep watch over engineering and communications. The textual content offers a lucid remedy of techniques and idea, which, supported via 250 labored difficulties and issues of solutions, makes the amount a worthy instructing reduction for the lecturer and an invaluable textual content for self university by way of the scholar. This revised and enlarged version features a new bankruptcy on two-port networks, extending the usefulness of the amount in all undergraduate electric and digital engineering classes.

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**Additional resources for Electric Circuit Theory. Applied Electricity and Electronics**

**Example text**

Iii) If i(h) = 1 0 " A at t = t u then 1-90 -40r 10 or 2 x HT = 1 5 e - 41 0 = e ' . Taking logs to base e In 2 - 10 In 10 = - 4 0 r 1? 558 sec. 2) yields a solution in two parts. 2) modified by replacing u(t) by zero. This part is the complementary function. 2) is the sum of the two parts. W e are now able to identify these two parts of the solution with the forced and natural responses already discussed. Just as the purpose of the complementary function is to provide arbitrary constants for accommodating different initial conditions, so the natural response accommodates the initial circuit conditions, bridging the gap between the actual response at some specific instant (usually t = 0) and the response dictated by the forcing function.

23 Thus, v2(t) •• L2L3 VL2 + L, diiW dt Also "i ( 0 U diifrt dt i>2(t)dt. FIELDS, CIRCUITS AND CIRCUIT PARAMETERS 35 Hence, v(t) dii(t) dt VL 2+ L 3 L 2L 3 l d / , ( 0 L2 + L 31 dt dt dt Thus, L2 + L3 Hence, inductances combine as resistances do, viz. those in series have an equivalent inductance equal to the sum of their inductances, whilst those in parallel have an equivalent inductance equal to the reciprocal of the sum of their reciprocal inductances. 21 Determine the equivalent capacitance of the three capacitors arranged as shown in the diagram of Fig.

H e n c e , 30 = Ri + ——— 10 + R2 or, 300 + 20iv 2 = WRi + RiR2. 25) Adding eqns. 25) gives: Ri = 2R2. 5Q. T o calculate the power in the load, we first determine the current in it. The generator feeds, in effect, two 10-Q resistors in series. 5 A . 2 x 10 T h e current through the load, 5 h = ~z tt—^ x ° - = 0-0918 A . 252 W. 5 W. 248 W. The price of matching both the generator and the load is, in this case, the loss of 9 0 % of the generated power. 10 D e t e r m i n e the resistance of and the current taken by a 100-W, 240-V lamp.