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*CARE TO SHARE. LEARN TO SUCCEED.*Recent Responses GATE 2015 : Electronics And Communication Engineering (Set 2) 0.6 mA

this is a typical wheatstone bridge. Consider the fact that for any current to pass through 1k resistor at the middle, there must be voltage difference at both sides.

Now, in a wheatstone bridge configuration, we have two voltage dividers installed. These in this case are namely 2K with 4K; and 3K with 6K. The difference of outputs of these voltage dividers will actually decide how much current will pass through the middle branch.

Although there is an external CURRENT source applied, it does not change the fact that due to whole circuit resistance, there will be some voltage appearing at the top and bottom of the wheatstone bridge.

We call that voltage as V

_{TH}. Now if we consider another wheatstone bridge with same resistors but an external VOLTAGE supply equal to V_{TH}, we can see that the ratio of resistors in each of the mentioned potential dividers actually output a voltage equal to V_{TH}/ 2.Thus, in the middle branch, there is no voltage difference, and thus no current flowing. Thus, the middle branch is non-existent.

And we can redraw the circuit without the middle branch of 1K resistor and the diode.

the new circuit will consist of following arrangement.

[2K + 4K] || [3K + 6K]

and thus

6K || 9K

to find the current flowing through 4K resistor, use the current division formula and the current obtained will be equal to 0.6mA

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