Q.
In the circuit given here, the points A, B and C are 70V, zero, 10 V respectively. Then
A) The point D will be at a potential of 60V
B) The point D will be at a potential of 20V
C) Currents in the path AD, DB and DC are in the ratio of 1:2:3
D) Currents in the path AD, DB and DC are in the ratio of 3 : 2 : 1
Explanation
(d) Applying Kirchhoff’s law at point D, we get
$I_{1} = I_{2} + I_{3}$
$\frac{V_{A} - V_{D}}{10} = \frac{V_{D} - 0}{20} + \frac{V_{D} - V_{C}}{30}$
$or\ \ \ 70 - V_{D} = \frac{V_{D}}{2} + \frac{V_{D} - 10}{3}$
$\Longrightarrow V_{D} = 40V\ $
$\Longrightarrow i_{1} = \frac{70 - 40}{10} = 3A\ $
$i_{2} = \frac{40 - 0}{20} = 2A\ $
$and\ i_{3} = \frac{40 - 10}{30} = 1A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.