In the network shown in fig. , points A, B, and C are at potentials of 70 V, 0, and 10V, respectively.

Options

A. the potential of point D is `40V`
B. The current in the sections `AD,DB,DC` are in the ratio `1:2:1`
C. The current in the sections `AD,DB,DC` are in the ratio `1:2:3`
D. the network drawn a total power of `200 W`

Answer

a,b,d

Explanation

Refer Let `V` be the potential of point `D`

Then `70 - V = 10I_(1)`….(i)
and `V - 0 = 20 I_(2) or V = 20 I_(2)`….(ii)
and `V - 10 = 30(I_(1) - I_(2))`….(iii)
From (ii) and (iii),
`20 I_(2) - 10 = 30 I_(1) - 30I_(2)`
`30I_(2) - 50I_(2) = -10`....(v)
On solving we get
`I_(1) = 3A,I_(2) = 2A, (I_(1) - I_(2)) = 1A`
and `V = 40 V`
`:. ` Total power conssumed
`= (3)^(2) xx 10 + (2)^(2) xx 20 + (1)^(2) xx 30` M
`= 90 + 80 + 30 = 200 W`