Question 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 (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 /n(1 point) = For the given circuit, Solve for Vo, Ip, Iy,, and Iz, for the following cases. Take Vg = 1 V, Vp = 0.25 V, Rp = 15 Q, and Vz =0.7V. a) R; = 8000, Ry = 2002 b) R, = 500, Ry = 5000 c) R; = 2002, Ry = 8002 d) Ry = 500, Ry, = 9500 VoV IZA ILA IpA a) b) | | °) | | q)

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For the given circuit, Solve for VO , ID , IL , and IZ for the...

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 Ω
b) R1=500Ω , RL=500Ω
c) R1=200Ω ,RL=800Ω
d) R1=50Ω ,RL=950 (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 Ω

(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

Transcribed Image Text: (1 point) = For the given circuit, Solve for Vo, Ip, Iy,, and Iz, for the following cases. Take Vg = 1 V, Vp = 0.25 V, Rp = 15 Q, and Vz =0.7V. a) R; = 8000, Ry = 2002 b) R, = 500, Ry = 5000 c) R; = 2002, Ry = 8002 d) Ry = 500, Ry, = 9500 VoV IZA ILA IpA a) b) | | °) | | q)

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Answer #1Q Pns _(n)rarr In fig(2) We show that the we have to Calculate the V_("th ") and Rth accross point AB and Ag(3) is theunin Ckt accross AB.V_("th ")=(R_(L)V_(S))/(R_(1)+R_(L))quadR_("th ")=R_(1)//∣R_(L)fig (1)rarr diode and zener diode will not Work simultanionely. becaule v_(z)=0.7v diode will turn on. and that's why. max voltage accross zener diode is 0.25V.(a) R_(1)=800quadR_(2)=200 OmegaAg(3)V_("th ")=(200)/(1000)xx1=0.2VV_("th ") < V_(D) So diode and zenerI_(0)=0quad&quadI_(L)=(V_(S))/(R_(1)+R_(L))=(1)/(1000)=1mA,I_(2)=0,V_(0)=0.2vffg(B)(b) R_(1)=500 OmegaquadR_(L)=500 OmegaV_("th ")=(500)/(1000)xx1=.500V=0.5VV_("th ") > V_(D) So diode turn On. V_(0)=0.25V{:[I_(S)=(1-0.25)/(500)=1.5mAquadI_(L)=(V_(D))/(R_(L))=(0.25)/(500)],[I_(L)=0.5mA],[I_(2)=0quadI_(D)=I_(S)-I_(L)],[I_(D)=1.5-0.5=1mA]:}(c){:[R_(1)=200quadR_(2)=800],[V_("th ")=0.8v]:}fig(5)V_("th ") > V_(2) & V_("th ") > V_(D) (But diode furn off zencr diode)V_(AB)=V_(D)=0.25VquadV_(0)=0.25VSo I_(s)=(1-0.25)/(200)=3.75mAquadI_(L)=(0.25)/(800)=0.3125mAI_(2)=0mAquadI_(D)=I_(S)-I_(L)=3.75mA-0.3125=3.4375mA(d) R_(1)=50 OmegaquadR_(L)=900 OmegaV_(H)=(9Omega0)/(1000)xx1=0.95VSo V_("th ") > V_(D)rarr diode Will on.{:[V_(AB)=V_(D)=0.25V] ... See the full answer