Home / Engineering & CS /For the given circuit, Solve for VO , ID , IL , and IZ for thefollowing cases. Take VS=1 V , VD=0.25 V , RD=15Ω , and VZ=0.7 V.a) R1=800 Ω , RL=200 Ωb) R1=500Ω , RL=500Ωc) R1=200Ω ,RL=800Ωd) R1=50Ω ,RL=950Ω (1 point) Ov: 141 ŠR lo vo @ 0 For the given circuit, Solve for Vo, ID, IL, and Iz for the following cases. Take Vs = 1V, Vp = 0.25 V, Rp = 15 12, and V2 = 0.7 V. a) Ri = 800 12, R = 200 12 b) R1 = 500 12, RL = 500 12 c) R1 = 200 S2, RL = 800 32 d) R1 = 502, RL = 950 22 VOV Iz A ILA Ip A .001

Question For the given circuit, Solve for VO , ID , IL , and IZ for thefollowing cases. Take VS=1 V , VD=0.25 V , RD=15Ω , and VZ=0.7 V.a) R1=800 Ω , RL=200 Ωb) R1=500Ω , RL=500Ωc) R1=200Ω ,RL=800Ωd) R1=50Ω ,RL=950Ω (1 point) Ov: 141 ŠR lo vo @ 0 For the given circuit, Solve for Vo, ID, IL, and Iz for the following cases. Take Vs = 1V, Vp = 0.25 V, Rp = 15 12, and V2 = 0.7 V. a) Ri = 800 12, R = 200 12 b) R1 = 500 12, RL = 500 12 c) R1 = 200 S2, RL = 800 32 d) R1 = 502, RL = 950 22 VOV Iz A ILA Ip A .001

CEGT86 The Asker · Electrical Engineering

For the given circuit, Solve for VO , ID , IL , and IZ for the following cases. Take VS=1 V , VD=0.25 V , RD=15Ω , and VZ=0.7 V .a) R1=800 Ω , RL=200 Ω
b) R1=500Ω , RL=500 Ω
c) R1=200Ω , RL=800Ω
d) R1=50Ω , RL=950Ω

Transcribed Image Text: (1 point) Ov: 141 ŠR lo vo @ 0 For the given circuit, Solve for Vo, ID, IL, and Iz for the following cases. Take Vs = 1V, Vp = 0.25 V, Rp = 15 12, and V2 = 0.7 V. a) Ri = 800 12, R = 200 12 b) R1 = 500 12, RL = 500 12 c) R1 = 200 S2, RL = 800 32 d) R1 = 502, RL = 950 22 VOV Iz A ILA Ip A .001
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Transcribed Image Text: (1 point) Ov: 141 ŠR lo vo @ 0 For the given circuit, Solve for Vo, ID, IL, and Iz for the following cases. Take Vs = 1V, Vp = 0.25 V, Rp = 15 12, and V2 = 0.7 V. a) Ri = 800 12, R = 200 12 b) R1 = 500 12, RL = 500 12 c) R1 = 200 S2, RL = 800 32 d) R1 = 502, RL = 950 22 VOV Iz A ILA Ip A .001
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:. when both dide open ckt thenV_(0)=(R_(L))/(R_(1)+R_(L))xxV_(S)=(200)/(200+800)xx1=0.2Volt:' It is less than both v_(2) and v_(D):. both zener diode and the dibde will be off. and v_(0)=0.2v_(0)t=quadn_^(Ans) i^(˙)_(D)=DA quad Pns i_(z)=OA quad Ans{:[i_(L)=(V_(S))/(R_(1)+R_(L))=(1)/(800+200)A=1mA],[=>l^(˙)_(L)=1mAquad" Ans "]:}(b) R_(1)=500 Omega,R_(L)=500 Omega when diodes are not persent:.v_(0)=v_(S)xx(500)/(500+500)=0.5" volt ":'quadv_(3) < v_(0) < v_(2):. Diode will be ON and zener diode will be off.:. Circuit will beApply KIL{:[:.V_(S)=R_(1)(I_(L)+I_(D))+R_(L)I_(L)],[=>1=500(I_(L)+I_(D))+500I_(L)],[=>1000I_(L)+500I_(D)=1]:}Agein apply KvL in loop-(ii){:[I_(L)R_(L)=I_(D)R_(D)+0.25],[=>500I_(L)-500I_(D)=0.25]:}From equection (i) and (ii)=>I_(L)=0.83mA AnsI_(D)=0.33mA Ans:' Zenerdiode is open ckt:.I_(2)=OA AnsNow: V_(D)=R_(L)I_(L)=500 xx0.83 xx10^(-3)V=>V_(0)=0.415V Ans(C) As givenR_(1)=200 Omega,R_(L)=800 Omegawhen both diode open thenV_(0)=(R_(L))/(R_(1)+R_(L))xxV_(S)=(800)/(6000)xx1=0.8V:.quadV_(0) > V_(2) > V_(D):. In this lose both drode will be ONApsly kVL loop -(D){:[:.V_(S)-R_(1)(I_(2)+I_(L)+I_(D))-gamma_(d)I_(2)-V_(2)=0],[=>1-200(I_(2)+I_(L)+I_(D))-15I_(2)-0.7=0],[=>215I_(2)+200I_(L)+200I_(D)=0.3]:}Again apply kve in loop-(1){:[V_(2)+r_(d)I_(2)=I_(L)R_(L)],[=>0.7+15I_(2)=800I_(L)],[=>15I_(L)-800I_(L)=-0.7]:}Agiin apply KVL at loop-(III){:[I_(L)R_(L)=r_(d)I_(D)+V_(D)],[=>800I_(L)=15I_(D)+0.25],[=>800I_(L)-15I_(D)=0.25]:}By solving whove eqn (is, (ii) and (iii){:[I_(2)=-14mA],[I_(L)=0.61mA],[I_(D)=15.9mA]:}As above Result zenerdiode loment is negative:. The zener diode mumt be off-state.:. Hew ... See the full answer

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