Question

Part 2 Maximum Power Transfer (Validating the Condition R = Rth) (a) Construct the network of Fig. 11.7 and set the potentiometer to 50 2. Measure the voltage across R as you vary Rthrough the following values: 50, 100, 200, 300, 330, 400, 600, 800, and 1000 12. Be sure to set the resistance with the ohmmeter section of your meter be fore each reading. Remember to turn off the dc supply and disconnect one terminal of the po tentiometer when setting the resistance level. Complete Table 11.3 and plot P versus R on Graph 11.1. R = 330 0 kn E8V FIG. 11.7

Answer 1

Theoritical Explanation RL= 3201 I = pt PL E = SVT VL = 'I h = /RL R+ R 1) - Pa I / ² RL Re: Son Vea L.05 V ; PL = 0.02205 W VL = 1.86 V. ; R = 0.0346 W VL = 3.02 V ; P=0.0456W, Ve= 3.81V ; P = 0.0484 W I = 0.021 A i (1) RL = LOOR _ I = 0.0.186 (L) PL = 2002 I = 0.0151 A ; (1) R = 300k 1 I = 0.0127A ; w R = 330 el I = 0.0121A į (1) R = 4002 . I = 0.01LA : (vill R = 6oon I = 0.0086A - RL = 8002 I=0.0071A; (1x) R = 10002 I =0.006 A ; liveAv 1 = 0.048 5W/ V = 4.38v s p = 0.0480 W Ves 5.16V : Pe=0.0444 w U = 5.66 h : PE 0-040W V2 = 601 VW P=0.036W When (RL = R. on TRL = R1 results in maximum output Pone which is the oritically and experimantly true .

CASO CASOS Rimw 0.1 0.2 0.3 0.4 0.5 0.6 067 os on Rh

(6) R = R1 = 3302 Theoritically - for Pmax Plot also it is true Ve = t = 4V @ maximum Power form Plot also it is True