Q. A 3 V battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I, in the circuit will be
A) 1A
B) 1.5A
C) 2A
D) $\frac{1}{3}A$
Explanation
(b) Resistance in the arms AC and BC are in series,
∴ R’=3+3=6$\Omega$
Now, R’ and 3$\Omega$ are in parallel,
$\therefore\ \ R_{eq} = \frac{6 \times 3}{6 + 3} = 2\Omega$
Now, V=IR
$\Longrightarrow \ \ I = \frac{3}{2} = 1.5A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.