Question

- WHEEL of (mis) FORTUNE 4 SETTINGS. VOLTAG AT THE WHEEL BELOW CAN RUTATE TO STOP AT EACH OF DETERMINE THE SAW WHICH PROVIDES 77+ morom POINT 'X' AND STATE WHAT ITS VALVE IS. DIODES NEED 0.7V TO TURN ON BE Precise. 1 +6V O WHEEL VOLTAGE m 1+4vh ov +2V X max = Posios

Answer 1

Position 1 (+6 0.7 M T+4h +4 HWwk Forward bias Let assume all diodes are in then find Voltage at node x Apply KCL at node x Va-2.7 V2-3.3 2 Tī Va-6.7 Va-7.3 < + =0 +2 T Valt+1+1+5) = 2;#49.3 +6.7 + 7 Va (3) = 15 Va = 5V por VX = 5V + Diodes connected to 4V & 6V are in Reverse bias Dides connected to QV 48V are in Forward bias

Hence new circuit from above conclusion + 6 xw 0.75 +2 .+ 2 2 2 Now apply KCL at nodex, by considering diodes are in FB Vz-2.7 + V2-703 =0 Vz(+1)-2.747,3 - Va = 10/ V2 = 5V Hence for position 1 Voltage at nodex is 5 Xi = 5 V

Positions +4) 4.0.7 KMM 2 -AWE tto. &2 (+8) Apply KCL at node x by considering all diodes are in FB Vx-8.7 Va-1.3, VX-407 - + V Vx-5,3-0 - +- Ve (+1 +1+3) = 8:* +1+3 +42+ + 533 VaR (3) = 13 Vze = 13 = 4.33V for Va=4.33V Diodes connected to 4V, 6V are in Forward bias and others are in Reverse bias

Hence new circuit from above conclusion 1+4 * 0.7 si Voo Now apply KCL at nodex, by considering diodes are En FB VX-407 + VQ-5,3 = 0 1 2 VR (1+1%) = 4•7+5:30 Va = 4.9V Hence for position 2 Voltage at nodex is 5 X2=4.9V

Position 3 +2 $ 0.7 + - ttoo (+6 Now apply KCL at nodex, by considering diodes are in FB Væ-6.7 V8-7.3 Vee-2.7 V6-3,3 -0. etī 2 2 va (3 +1+1+a) = 6,7 +7.3+267 + 3,3 va (3) = 15 - Va=5v for Va=5v diodes connected to 8V, ar are in FB and others are in RB

Hence new circuit from above conclusion +2 x NEW +0.7 +6 Now apply KCL at nodex, by considering diodes are in FB Va-7.3 Vxe-2.7 Va(1+1) = 7.3 +2.7 Va(2) = lov Voe=5v Hence for position Voltage at nodex is 5 Xg=5v

Position 4 + 8 0.7 x 24- +6Hm Amt ttoo 0.7 2 0.755 (+4) Apply KCL at nodex by considering all diodes are in F.B Vx-407 va 5.3 Va-8.7 Væ-1.3 +25.3 + — + 20 - 17 v ( +1+1+1)= 4 + + 5.3+8.4 Ve (3) = 14 Va = 11 / 2 = 5.667V for Vx=5.667V Diodes connected to 4V, 6V are in Forward bias and others are in Reverse bias

+8 +0. 2 0.7* (+4) How apply KCL at nodex, by considering diodes are in FB Ve-4.7 Vor-5,3 27 + 122-5.3 Va (4+) = 4*+503 VX (15) = 7.65 V x = 7.65 = 5.1V Hence for position 4 voltage at nodex is 5 X4 = 5.1 V

Hence Xmax = 5.1V fou poration 4