question_answer

In the circuit given here, the points A, B and C are 70 V, zero, 10 V respectively. Then

A)  the point D will be at a potential of 60 V

B)  the point D will be at a potential of 20 V

C)  currents in the paths AD, DB and DC are in the ratio of 1 : 2 : 3

D)  currents in the paths AB, DB and DC are in the ratio of 3 : 2 : 1

Correct Answer: C

Solution :

Applying Kirchhoffs 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 \[70-{{V}_{D}}=\frac{{{V}_{D}}}{2}+\frac{{{V}_{D}}-10}{3}\] \[\Rightarrow \]\[{{V}_{D}}=40V\]\[\Rightarrow \]\[{{i}_{1}}=\frac{70-40}{10}=3A\] \[{{i}_{2}}=\frac{40-0}{20}=2A\]and\[{{i}_{3}}=\frac{40-10}{30}=1A\]