Using Kirchoff's rules determine the value of unknown resistance R in the circuit shown in figure, so that no current flows through `4 Omega` resistance. Also find the potential difference between A and D.
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The current I flows in the circuit. Applying Kirchhoff's second rule to loop AFEBA, we have `1 xx I + 1 xx I = - 6 +9 =3 or I=1.5A` From the loop BEDCB, we have `IR=6 -3 = 3 or R = 3/1.5 = 2 Omega` `V_(AD) = (1+1) xx I = 2xx1.5 = 3 V`
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