Q. Two cells having the internal resistance 0.2$\ \Omega$ and 0.4$\ \Omega$ are connected in parallel. The voltage across the battery terminal is 1.5V. the emf of first cell is 1.2V. the emf of second cell is
A) 2.7 V
B) 2.1 V
C) 3 V
D) 4.2V
Explanation
(a) Given that, the resultant voltage across the battery terminal=1.5V
Let I be the current in the circuit then total resistance =0.6$\Omega$
Hence, $V = IR\ $
$\Longrightarrow 1.5 = I \times 0.6$
$\Longrightarrow I = \frac{1.5}{0.6} \Longrightarrow I = \frac{5}{2}A$
Now, applying Kirchhoff’s second law in the circuit
0.4I + 0.2I + 1.2 - E=0
$0.6\ \times \frac{5}{2} + 1.2 = E\ $
$\Longrightarrow \ \ \ E = 2.7V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.