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Determine the equivalent resistance of networks shown in figure. |
Answer» Solution : EQUIVALENT resistance of one loop (parallel connection of 2 `Omega and 4 Omega `) `R. = (2 xx 4)/(2 + 4) = (8)/(6) = (4)/(3) Omega` Now, given network can be SHOWN as follow, `rArr ` Equivalent resistance of given network will be  R = 4 R. = `4 xx (4)/(3)` `= (16)/(3) Omega = 5.3 Omega` (ii) Here FIVE resistances each ofR `Omega` are connected in series and so that equivatent resistance would be, `R_(S) = R + R + R ` + R +R `THEREFORE R_(S) = 5 R Omega` |
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