For the network shown in figure, points A, B and C are at potentials of 70 V, zero and 10 V respectively

Options

A. point D is at a potential of 40 V
B. the currents in the sections AD, DB and DC are in the ratio `4: 3: 2`
C. the current in the sections AD, DB and DC are in the ratio `1: 2: 3`
D. the network draws a total power of 100 W

Answer

A

Explanation

Consider the current distributions in the circuit as shown below

By `KVL, V_(A)-l_(1)10-l_(2)20-V_(B)=0`
`implies l_(1)10+l_(2)20=70`
`V_(A)-l_(1)10-l_(3)30-V_(C )=0`
`implies " " l_(1)10+l_(3)30=60`
Also, `l_(3)=l_(1)-l_(2)`
So, Eq. (ii) becomes
`l_(1)10+30(l_(1)-l_(2))=60`
`l_(1)40-30 l_(2)=60`
Solving Eqs. (i) and (iii), we get
`l_(2)=2 A, l_(1) =3A, l_(3)=1A`
Potential drop across branch AD,
` V_(A)-V_(D)=l_(1)xx10 implies 70-V_(D)=30 implies V_(D)=40 V`