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saubhik
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In the attachment (also given at bottom for your convenience) you can see the circuit diagram. We all know that the 5 ohm resistor is short-circuited so no current flows through the resistor. But if we apply Kirchoff's loop law with the following symbols:
from loop AEFBA,
Can anyone please explain me where I am going wrong ?
Also if we apply loop law to BDCAB,
Where is the fallacy here? (i mean, where am i wrong ? )
Thank you.
[PLAIN]http://img594.imageshack.us/img594/7609/circuit.jpg
i = current through DBAC from positive terminal of battery
i1 = current through CD from junction C
i-i1 = current through CEFD from E
we get,i1 = current through CD from junction C
i-i1 = current through CEFD from E
from loop AEFBA,
5(i-i1) = 5 or i-i1 = 1
from loop CEFDC,5(i-i1) = 0 or i-i1 = 0
Thus ambiguity arises.Can anyone please explain me where I am going wrong ?
Also if we apply loop law to BDCAB,
5 = 0
Where is the fallacy here? (i mean, where am i wrong ? )
Thank you.
[PLAIN]http://img594.imageshack.us/img594/7609/circuit.jpg
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