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The_solutions_to_CHT2s_homework

# The_solutions_to_CHT2s_homework - Chapter 2 Solution 8 8A...

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Unformatted text preview: Chapter 2, Solution 8. 8A At node a, At node 6, At node (1, Chapter 2, Solution 9. AtA, 2+12=f1 —> AtBj 122f2+14 —> AtC, 1424+:3 —> 9Ad 8=12+i1 —I'- 9=8+i2 —I- 9 12+l3 —I' i! = — 4A 1;: 1A i§=-3A Chapter 2, Solution 14. Formeshl, —V4+2+5=0 —> V4=7V FormeshZ, +4+V3+V4=0 —> V3=—4—?=-HV Formesh3, —3+pf_,—V3=0 —> V1:V3+3=—3If Formeshﬁlj —V£—V2—2=0 —> sz—VI—226V Thus, VI =—8V, V2 = 6V, V} =—HV, V4 = 717 Chapter 2, Solution 15. Forlnnp 1, —12+v+2=0, v=10V For loop 2, —2 + 8 + 3i}; =0, ix = —2A Chapter 2, Solution 19. Applying KVL around the loop, we obtain -12+10-(-3)+3i=0—I- i=—2A Power dissipated by the resistor: pm =iER=4(3)=m Power supplied by the sources: Plzv = 12 01—2)) 2 ﬂ va = 10 ({4}) = M Psv = (-3)(-2) = ﬂ Chapter 2, Solution 20. Applying KVL around the loop, -36+4in+5ig=0 —F iﬂ=4A Chapter 2, Solution 27. Using voltage division, P; : i(16V) : 6.4 V 4 + 16 Chapter 2, Solution 24. _ V; (a) IO_R1+R2 {1V5 R3R4 R1+ R2 R3 + R4 V0 : ﬁll” (R3HR4) : 5 _ _ “33134 Vs _(R1 +R2IR3 +34) (1)) IfR1=R2=R3=R4=R3 VB vs i5 2 :Ele—b (1:40 2R 4 — Chapter 2, Solution 29. Reg]: 1 + 1110+ 1H2) = 1 + 1rr(1+ 213)=1+ 11f5f3=1-625f2 Chapter 2, Solution 30. We start by combining the 6-ohm resistor with the 2-ohm one. We then end up with an S-ohm resistor in parallel with a 2-ohrn resistor. (2x8)/(2+8) = 1.6 Q This is in series with the 6-ohm resistor which gives us, Req = 6+1.6 = 7.6 o. ...
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The_solutions_to_CHT2s_homework - Chapter 2 Solution 8 8A...

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