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In the circuit shown, if a conducting wire is connected between point A and B, the current in this wire will |
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Answer» flow from `A` to `B` Resistance `4 Omega` and `4 Omega` are connected in series, so their effective resistance, `R' = 4 + 4 = 8 Omega` Similarly, `1 Omega` and `3 Omega` are in series, `R" = 1 + 3 = 4 Omega` Now `R'` and `R"` will be in parallel, hence effective resistance `R = (R' xx R'')/(R' + R'') = (8 xx 4)/(8 + 4) = (32)/(12) = (8)/(3) Omega` Current through the circuit, from Ohm's LAW `i= (V)/(R ) = (3V)/(8) A` Let current `i_(1)` and `i_(2)` flow in the branches as shown `8 i_(1) = 4 i_(2)` `i_(2) = 2 i_(1)` .....(i) Also `i= i_(1) + 2 i_(1)` `i_(1) = (V)/(8) A` and `i_(2) = (V)/(4) A` Potential DROP at `A, V_(A) = 4 xx i_(1) = (4 V)/(8) = (V)/(2)` Potential drop at `B, V_(B) = 1 xx i_(2)= 1 xx (V)/(4) = (V)/(4)` Since, drop of potential is greater in `4 Omega` resistance. It will be at lower potential than `B`, hence, on connecting wire between points `A` and `B`, the current will flow from `B` to `A`. 
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