Q. In the figure, current through the $3\Omega$ resistor is $0.8\ ampere,$ then potential drop through $4\ \Omega$ resistor is
A) $9.6\ V$
B) $2.6\ V$
C) $4.8\ V$
D) $1.2\ V$
Explanation
(c) Current through $6\Omega$ resistance in parallel with $3\Omega$ resistance $= 0.4\ A$
So total current $= 0.8 + 0.4 = 1.2\ A$
Potential drop across $4\Omega = 1.2 \times 4 = 4.8V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.