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Determine the equivalent resistance of networks shown in Fig.(i) and (ii). |
Answer» Solution : (i) The circuit contains four identical elements joined inseries. One such element is shown in Fig.and hastwo branches, having resistance of 2 `Omega (i.e., 1 Omega +2 Omega)`and 4 12 (i.e., 212 +212) respectively, joined in PARALLEL.` therefore ` Resistance of the element`R = (2 xx 4)/(2 + 4) = 8/6= 4/3 Omega` For whole series net equivalent resistance = `4R = 4 xx4/3 =16/3Omega` (ii) The combination shown has five RESISTANCES all joined in series. The total resistance is,thus, 5R. |
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