Q. The current in a simple series circuit is 5.0.A. when an additional resistance of 2.0$\ \Omega$ is inserted, the current drops to 4.0 A. the original resistance of the circuit in ohm was
A) 1.25
B) 8
C) 10
D) 20
Explanation
(b) Let the original resistance is R $\Omega$.
∴ $V = IR$
$\ \ \ V = 5 \times R = 5R\ \ \ \ \ldots$(i)
When 2 $\Omega$ resistance is inserted, then total resistance
=(R+2)$\ \Omega$
∴ $V\ = \ I’\ (R + 2) = 4(R + 2)$ …(ii)
From Eqs. (i) and (ii), we get
$5R\ = \ 4(R + 2)\ \ $
∴ R=8 $\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.