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Catégorie :Category: nCreator TI-Nspire
Auteur Author: tbo3199
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 3.73 Ko KB
Mis en ligne Uploaded: 09/09/2025 - 15:54:30
Uploadeur Uploader: tbo3199 (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : https://tipla.net/a4836412
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 3.73 Ko KB
Mis en ligne Uploaded: 09/09/2025 - 15:54:30
Uploadeur Uploader: tbo3199 (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : https://tipla.net/a4836412
Description
Fichier Nspire généré sur TI-Planet.org.
Compatible OS 3.0 et ultérieurs.
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1 pwr sup by dep source is 2ma r1 4.7k id 5Vx r2 1 r3 3 ------- --r2----- | | | | is r1 id r3 | | | | -------------------- goal: pid=pr2+pr3= id^2*r2+id^2*r3=id^2(r2+r3); solve id: id=5vx vx=is*r1; id=5vx; pid=id^2*(r2+r3) = 8.83kw 2 thev eq circuit where rL connected and power dissp in RL --------- |----------- |--rth--| | / vs r1 r2 vth rl vs r1 r2 | thev1 |-vab-| | thevfin | | -rl- | r3 r4 ------ | r3/r4 ------------ ------- ir13=vs/r1+r3 ir24=vs/r2+r4 vth=voc=vab=|v3-v4| v4=i4*r4=vb v3=i3*r3=va rth=(r1|r3)+(r2|r4) il=vth/rth+rl pL=il^2*rl=ans 3 determine thev eq between a-b, determine the value of the resistor that absorbs max power when connected between a-b what will be the max power absorbed. vth=v1=i1r1 a--|---v1---|----| r1 r2 id | | | b-------------- vth w kvl: vs-i1r1+r2(-i1-i2)=0. vs-i1r1-i1r2-i2r2=0. vs+i1(-r1-r2-0.02*r1*r2)=0. i1=-vs/(-r1-r2-0.02*r1*r2) vs+i1(-r1-r2) -i2r2=0. vs+i1(-r1-r2)-(0.02*i1r1*r2) =0. -a---*---vs---* it= vs/rt; rth=vth/isc; isc=in=it | | redraw w vth rth and rL; vL=vth/2; | r2 pmaxrL= vl^2/rl; | | -b---*-------- 4 calculate power supplied each source, choose easiest method NODAL -r5-*---*--r6-- KCL @ a: ir5=v1-va/r1+r5; ir2 =va-0/r2->va/r2 | | | | ir5+i1+i2-ir2=0 -> ir5+i1+i2=ir2 r1 r2 r3 r4 v1/r1+r5 + -va/r1+r5 + i1+i2=va/r2- -va/r1+r5 | | | | v1/r1+r5 + i1 +i2= va (1/r2+ 1/r1+r5) v1 | ^i1 ^i2 va=(v1/r1+r5 +i1 +i2)/(1/r2 + 1/r1+r5) --------------- pv1=ir5*v1=(v1-va/r1+r5)*v1 pi1=i1*vi1= i1*(va+i1r3) pi2= i2*vi2= i2*(va+i2(r4+r6)) 5 contribution of power supplied by each source to the load; superposition ---r1---*---r3--- start w i1; short b2; calculate rt=r1+rL||r3 | i1 | i2 | ir2r3=it=i1=v1/rt; vr2r3= r2||r3 * i1 b1 r2(rL) b2 Prl=Pr2= (vr2)^2/r2; short b1 and solve i2 |-------|-------| rt=(r1||r2)+r3; it= vb2/rt=ir1+ir2 =i3; vr12= it*r12 =vL; pr2= (vr12)^2/r2; pvr23= vr23^2/rt; pvr12=vr12^2/rt; 6 thev eq circ across open nodes. determine value of res connec btwn the open nodes max power from source. that power too. r1=2 v1=5v, Vd=0.1V2, r3=3.3 ---r1--(-5v+)--r2----o+ | | vd r3 -------------------o- vth=voc=i*r3; rth=vth/isc; kvl: 0.1V2-i*r1+vs-i*r2-i*r3=0; -vs+0.1*v2+I(-r1-r2-r3)=0-vs I= -vs/(0.1*r3-r1-r2-r3) ----r1--(-vs+)--r2---o | thev | ------------------o rt=r1+r2; it= vs/rt; rth=vth/isc; |-------rth-----| vth thev final rl -------------- max power across rl; Prl= (vth/2)^2/rl; hw1 use kvl to obtain a numerical value for the current labeled i in each circ. -(+v2-)-r1--| a) -r1-(-v2+)-(-v3+)-r2-r3-- | (-v3+) | | v1 | (-v1+) v4 ------r2----- |-----r4----------v5----| a) v1-v2-r1*i+v3-r2*i=0; b)-v1-r1*i+v2+v3-r2*i-r3*i-v3+v5-r4*i v1-v2+v3=r1*i+r2*i-> i(r1+r2); -v1+v2+v3-v4+v5=i(r1+r2+r3+r4) i=(v1-v2+v3)/(r1+r2); i=(-v1+v2+v3-v4+v5)/(r1+r2+r3+r4); Made with nCreator - tiplanet.org
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Compatible OS 3.0 et ultérieurs.
<<
1 pwr sup by dep source is 2ma r1 4.7k id 5Vx r2 1 r3 3 ------- --r2----- | | | | is r1 id r3 | | | | -------------------- goal: pid=pr2+pr3= id^2*r2+id^2*r3=id^2(r2+r3); solve id: id=5vx vx=is*r1; id=5vx; pid=id^2*(r2+r3) = 8.83kw 2 thev eq circuit where rL connected and power dissp in RL --------- |----------- |--rth--| | / vs r1 r2 vth rl vs r1 r2 | thev1 |-vab-| | thevfin | | -rl- | r3 r4 ------ | r3/r4 ------------ ------- ir13=vs/r1+r3 ir24=vs/r2+r4 vth=voc=vab=|v3-v4| v4=i4*r4=vb v3=i3*r3=va rth=(r1|r3)+(r2|r4) il=vth/rth+rl pL=il^2*rl=ans 3 determine thev eq between a-b, determine the value of the resistor that absorbs max power when connected between a-b what will be the max power absorbed. vth=v1=i1r1 a--|---v1---|----| r1 r2 id | | | b-------------- vth w kvl: vs-i1r1+r2(-i1-i2)=0. vs-i1r1-i1r2-i2r2=0. vs+i1(-r1-r2-0.02*r1*r2)=0. i1=-vs/(-r1-r2-0.02*r1*r2) vs+i1(-r1-r2) -i2r2=0. vs+i1(-r1-r2)-(0.02*i1r1*r2) =0. -a---*---vs---* it= vs/rt; rth=vth/isc; isc=in=it | | redraw w vth rth and rL; vL=vth/2; | r2 pmaxrL= vl^2/rl; | | -b---*-------- 4 calculate power supplied each source, choose easiest method NODAL -r5-*---*--r6-- KCL @ a: ir5=v1-va/r1+r5; ir2 =va-0/r2->va/r2 | | | | ir5+i1+i2-ir2=0 -> ir5+i1+i2=ir2 r1 r2 r3 r4 v1/r1+r5 + -va/r1+r5 + i1+i2=va/r2- -va/r1+r5 | | | | v1/r1+r5 + i1 +i2= va (1/r2+ 1/r1+r5) v1 | ^i1 ^i2 va=(v1/r1+r5 +i1 +i2)/(1/r2 + 1/r1+r5) --------------- pv1=ir5*v1=(v1-va/r1+r5)*v1 pi1=i1*vi1= i1*(va+i1r3) pi2= i2*vi2= i2*(va+i2(r4+r6)) 5 contribution of power supplied by each source to the load; superposition ---r1---*---r3--- start w i1; short b2; calculate rt=r1+rL||r3 | i1 | i2 | ir2r3=it=i1=v1/rt; vr2r3= r2||r3 * i1 b1 r2(rL) b2 Prl=Pr2= (vr2)^2/r2; short b1 and solve i2 |-------|-------| rt=(r1||r2)+r3; it= vb2/rt=ir1+ir2 =i3; vr12= it*r12 =vL; pr2= (vr12)^2/r2; pvr23= vr23^2/rt; pvr12=vr12^2/rt; 6 thev eq circ across open nodes. determine value of res connec btwn the open nodes max power from source. that power too. r1=2 v1=5v, Vd=0.1V2, r3=3.3 ---r1--(-5v+)--r2----o+ | | vd r3 -------------------o- vth=voc=i*r3; rth=vth/isc; kvl: 0.1V2-i*r1+vs-i*r2-i*r3=0; -vs+0.1*v2+I(-r1-r2-r3)=0-vs I= -vs/(0.1*r3-r1-r2-r3) ----r1--(-vs+)--r2---o | thev | ------------------o rt=r1+r2; it= vs/rt; rth=vth/isc; |-------rth-----| vth thev final rl -------------- max power across rl; Prl= (vth/2)^2/rl; hw1 use kvl to obtain a numerical value for the current labeled i in each circ. -(+v2-)-r1--| a) -r1-(-v2+)-(-v3+)-r2-r3-- | (-v3+) | | v1 | (-v1+) v4 ------r2----- |-----r4----------v5----| a) v1-v2-r1*i+v3-r2*i=0; b)-v1-r1*i+v2+v3-r2*i-r3*i-v3+v5-r4*i v1-v2+v3=r1*i+r2*i-> i(r1+r2); -v1+v2+v3-v4+v5=i(r1+r2+r3+r4) i=(v1-v2+v3)/(r1+r2); i=(-v1+v2+v3-v4+v5)/(r1+r2+r3+r4); Made with nCreator - tiplanet.org
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