MHT CET 2021 Physics Solved Paper

(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