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Find equivalent resistance of the network in fig. Between points (i) A and B and (ii) A and C. |
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Answer» (i) The `10Omega` and `30Omega` resistors are connected in parallel between points A and B. The equivalent resistance between A and B is `R_(1)=(10xx30)/(10+30)ohm=7.5Omega` (ii) The resistance `R_(1)` is connected in series with resistor of `7.5Omega` hence the equivalent resistance between point A and C is `R_(2)=(R_(1)+7.5)ohm=(7.5+7.5)ohm=15Omega`. |
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