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

Options

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 AD,DB and DC are in the ratio `3:2:1`.

Explanation

Applying Kirchhoffs law at point D, we get
`I_(1)=I_(2)+I_(3)`
`(V_(A)-V_(D))/(10)=(V_(D)-V_(B))/(20)+(V_(D)-V_(C))/(30)`
`(70-V_(D))/(10)(V_(D)-0)/(20)+(V_(D)-10)/(30)`
`70-V_(D)=(V_(D))/(2)+(V_(D)-10)/(3)`
`6(70-V_(D))=3V_(D)+2(V_(D)-10)`
`420-6V^(D)=3V_(D)+2V_(D)-20`
`11V_(D)=440`
`V_(D)=40 V `
`I_(1)=(V_(1)-V_(D))/(10)=(70-40)/(10)=3A`
`I_(2)=(V_(D)-V_(B))/(20)=(40-0)/(20)=2A`
`I_(3)=(V_(D)-V_(C))/(30)=(40-10)/(30)=1A`
`**` None of the given options is correct.