Chapter 28135P28.81201234Veq==+++FHGIKJIRIAAAAρρρρAAAA, or IAAAAρA=+++12011111234Vafej∆VIAAAAAA222111112029 51234==+++=ρAVVafej.P28.9If we turn the given diagram on its side, we find that it is the same as figure(a). The 20 0. Ωand 5 00.Ωresistors are in series, so the first reduction isshown in (b). In addition, since the 10 0. Ω, 5 00.Ω, and 25 0. Ωresistors arethen in parallel, we can solve for their equivalent resistance as:Req=++=12 94110 015 00125 0....ΩΩΩΩch.This is shown in figure (c), which in turn reduces to the circuit shown infigure (d).Next, we work backwards through the diagrams applying IVR=∆and∆VIR=alternately to every resistor, real and equivalent. The 12 94.Ωresistor is connected across 25.0 V, so the current through the battery inevery diagram isIVR===∆Ω25 01 93..V12.94 A .In figure (c), this 1.93 A goes through the 2 94.Ωequivalent resistor to give apotential difference of:∆ΩVIR===1 932 945 68...AVafaf.From figure (b), we see that this potential difference is the same across ∆Vab,the 10 Ωresistor, and the 5 00.Ωresistor.(b)Therefore, ∆Vab=5 68.V.(a)Since the current through the 20 0
This is the end of the preview.
access the rest of the document.