(d) The current distribution in the circuit is shown below
Applying KVL in loop PQSRP, we get −5I−RI1+12=0⋅⋅⋅⋅⋅⋅⋅(i) Similarly, for loop SRUTS, we get r−I1R+7=0 ⇒ I1R=7⋅⋅⋅⋅⋅⋅⋅(ii) As per question, I−I1=0⇒I=I1⋅⋅⋅⋅⋅⋅⋅(iii) Solving Eqs. (i), (ii) and (iii), we get −5I−7+12=0⇒I=1A From Eq. (ii), R=7Ω