Kirchhoff's Laws & Measuring Instruments question from MHT CET 2026 (19 May, Shift 2)

Answer: (d)

1. Reduce from the right: $2 + 4 + 2 = 8\ \Omega$, in parallel with $8\ \Omega$ gives $4\ \Omega$.
2. Add $2 + 2$: $8\ \Omega$, in parallel with the first $8\ \Omega$ gives $4\ \Omega$.
3. Total $= 3 + 4 + 2 = 9\ \Omega$, so $I = \frac{9}{9} = 1$ A (this flows in $3\ \Omega$).
4. Equal $8\ \Omega$ splits: $1 \to 0.5$ A $\to 0.25$ A in the right branch containing $4\ \Omega$.
5. So the answer is (d) the $4\ \Omega$ resistance carries $0.25$ A.