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Why are we getting zero current in the branch EF of the previous example ? |
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Answer» Solution :Let us disconnect the points E and D of the circuit used in the previous example. Then the situation is as follows: Current i flowing from the positive terminal of the 12V battery will continue to flow in the `3Omega` resistance. A loop involving branch EF is OPEN so current will not flow through this branch. For rest of the circuit, current will be `i= (12)/(3+2+1)= 2A` Now we can see that `V_(F) - V_(E )= 6V` `V_(G) - V_(D) = i xx 3=6V`...(ii) Let potential of common points F and G be zero then: `V_(E) = -6V` `V_(D) = -6V` We have proved that the potential of points E and D are same when E is not CONNECTED with D. HENCE, if E and D are joined by a meal wire, then no current will flow through that wire. Now you can understand that the above question was based on the idea of a potentiometer. 
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