Question

Solve each practice problem. TYPE solutions in engineering notation

TYPE solutions in engineering notation.

Calculate the component voltages and branch currents for the circuit shown in Figure 6.40, along with the values of I, and Rr. 3. R3 2 kn R4 4.7 k R1 10 k Vs 26 V R5 3.3 k R2 3 kn FIGURE 6.40 Calculate the component currents and loop voltages for the circuit shown in Figure 6.42, along with the values of I and Rr 5. R1 220 A A R2 100 Vs 6 V R3 510 R4 470 B FIGURE 6.42 Calculate the component currents and loop voltages for the circuit shown in Figure 6.45, along with the values of Ir and Rr. 8. R1 5.6 kQ A R2 A 12 kn B R4 R3 2 k Vs 18 V 3.6 kn R5 7.5 k B R6 C 10 kn FIGURE 6.45 14. Determine the value of load voltage (V) for the circuit shown in Figure 6.46b. R R2 3 kf R 2.2 k 22 0 1 k R2 R4 R 300 R3 1.2 k 33 51 n Vs 12 V RL 1.5 k RL 39 R3 120 0 Vs 9 V (b) (a) FIGURE 6,46 17. Determine the value of load power (P) for the circuit shown in Figure 6.48a. R 120 100 RL 150 30 1 Vs 20 V Vs 9V- R. 100 R 150 R3 10 R2 200 15 n a) (b) FIGURE 6,48 wl ww w wl

Answer 1

luetion Le+I,T le tHe anch Cusrents des Leatively a IT VI Ry= 47k2 REE3.3k2 R23k Cuetent Me Circut a shoun above ->Let IT be the Tpta RT be the eguival ert desistance f tMe Citcurt nertioned >Bknaun that tn Padalle Qominotien, Voltoge aluayp Le Same Hhe CPamllel Qehnefed elements We get a Closs So by usiohn's Ca TTRR) Ans. 19K Ans Rst Ryt RE = keL at Nocle B,e t IT= It Ir= 2MA+2.6mA IT 4 6 MA Ans V= !okx 2 mA Veltage de acs Me R, degistey V, = R¢J, Voltage clet a cHess He R des iskr V= RI V= 3k2X 2mA Voltje dtep aca098 e Rs desiske, V3= RsIp V = 2kx 2:mA Voltage dach acns He Ry iste V=RI V= 417k2X2(m 5.2 V V4=12.22Ve AnS VoU tage dlacb acress he Rs desitt V5=RsI >V5= 3.3Kl X27MA V5=8.50 Vo Ans Har RTi R3,Ry Re ae n seics bm bindio So equivalaent o Hen Now TRE= loK S RE RatRyt Rs R. & R2 ave tn seies ombinatian go eguivalent RE;=R¢+R>R,=Bk9] GAEl=B >9a modiied Ciscut a soun telo wwWw

IT B I2 2vol4 T RE13KRE1D k2 New RE, RE ae n Pasalel Combination So equivalent TotalResisee Qieit isgny RTE RERE aftolvig,Ce RT 130 13+lo RT= 5.6521 k} Ane.re RErtRE RT 13 X1o