1.

In the given circuit, each resistance is r = 18.75Omega. The current(in A) in the resistance connected across A and B is '………..'A. .

Answer»


Solution :At y according to KIRCHHOFF's junction law,
`(y-x)/R + (y-x-100)/r + (y-50)/r + (y/r) + (y -50)/r = 0`
` 5y - 2x = 200`………….(i)

Similarly at x,
`i = (50-x)/r + (y-x)/r`……..(ii)
At x + 100,
`i = (x + 100 - 50)/r + (x+100-y)/r` .......(III)
From eqs. (ii) and (iii), we GET
`y - 2x = 50`........(iv)
From eqs. (i) and(iv), y = 37.5 V.
So, CURRENT through R is `37.5//18.75 = 2A`.


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