Q 12-03-035NEETJEE MainMedium
A $6$ V battery of negligible internal resistance is connected across $3\ \Omega$ in series with a parallel combination of $6\ \Omega$ and $3\ \Omega$. The current through the $6\ \Omega$ resistor is
Answer: (D) $0.4$ A
$6\ \Omega \parallel 3\ \Omega = 2\ \Omega$. Total $= 5\ \Omega$, so $I = \dfrac{6}{5} = 1.2$ A.
Voltage across the parallel pair $= 1.2 \times 2 = 2.4$ V, so the $6\ \Omega$ resistor carries $\dfrac{2.4}{6} = 0.4$ A.
Solution by Sreeraj P, M.Sc Physics