Q. Four identical resistors of 4 $\Omega$ each are joined in circuit as shown in figure. The cell $B$ has emf 2V and its internal resistance is negligible. The ammeter reading is
A) $\frac{3}{8}A$
B) $2A$
C) $\frac{1}{2}A$
D) $\frac{1}{8}A$
Explanation
(a) Here three resistance of $4\Omega$ each are connected in parallel so that their combined resistance $= \frac{4}{3}\Omega$. It is in series with ammeter, battery and last $4\Omega$ resistance.
$\therefore Net\ resistances\ R = \frac{4}{3} + 4 = \frac{16}{3}\Omega$
$\therefore$ Current in main circuit =ammeter reading
$i = \frac{E}{R} = \frac{2V}{\frac{16}{3}\Omega} = \frac{3}{8}A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.