30 - 3i2/8 - i2 - i2/2 = 0 or i2 = 2 A. From the value of i2, we now use Eqs. (2

Question

30 - 3i2/8 - i2 - i2/2 = 0 or i2 = 2 A. From the value of i2, we now use Eqs. (2.8.1) to (2.8.5) to obtain i1 = 3 A. i3 = 1 A. v1 = 24 V. v2 = 6 V, v3 = 6 V Find the currents and voltages in the shown in Fig. 2.28. Answer: v1 = 3 V, v2 = 2 V, v3 = 5 V, i1 = 1.5 A, i2 = 0.25 A, i3 = 1.25 A. Figure 2.28 For Practice Prob. 2.8. The need to combine resistors in series or in parallel occurs so frequently at it warrants special attention. The process of combining the resistors facilitated by combining two of them at a time. With this in mind, onsider the single-loop circuit of Fig. 2.29. The two resistors are in

Explanation / Answer

i3=i1-i2 now apply voltage law in 2 loops -(i1-i2)*4+3 +8*i2=0 for the other loop 5-2*i1-8*i2=0 solving above 2 we get i1=1.5A i2=.25A i3=1.5-.25=1.25A V1=I1*2=1.5*2=3V V2=.25*8=2V V3=1.25*4=5V

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