Q. Two resistors of resistances $200\ k\Omega$ and $1M\Omega$ respectively form a potential divider with outer junctions maintained at potentials of $+ 3V$ and $- 15V.$ Then, the potential at the junction between the resistors is

A) $+ 1\ V$
B) $- 0.6\ V$
C) $0\ V$
D) $- 12\ V$
Explanation

(c) Potential difference across $1\ M\Omega$ resistor is
$V_{P} - V_{B} = \frac{18V \times 1 \times 10^{6}\Omega}{(0.2 + 1) \times 10^{6}\Omega} = \frac{18V \times 1 \times 10^{6}\Omega}{1.2 \times 10^{6}\Omega} = 15V$
$V_{B} = - 15V$ [Given]
$\therefore V_{P} - V_{B} = 15V$ or $V_{P} = 15V + V_{B}$
$= 15V - 15V = 0V$
[image:3_image133.emf]
Potential difference across $200\ k\Omega$ resistor is
$V_{A} - V_{P} = \frac{18V \times 0.2 \times 10^{6}\Omega}{(0.2 + 1) \times 10^{6}\Omega}$
$= \frac{18V \times 0.2 \times 10^{6}\Omega}{1.2 \times 10^{6}\Omega} = 3V$
$V_{A} = + 3V$ [Given]
$\therefore V_{A} - V_{P} = 3V$ or $V_{P} = V_{A} - 3V$
$= + \ 3V - 3\ V = 0\ V$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.