Calculate the potentials of points `A, B,C` and `D` as shown in Fig. a. What would be the new potential values if connections of `6 V` battery are reversed as shown in fig b. All resistance are on ohm.

Options

A. `V_(B)=6V,V_(C)=9V,V_(D)=11V`
B. `V_(B)=11V,V_(C)=9V,V_(D)=6V`
C. `V_(B)=9V,V_(C)=11V,V_(D)=6V`
D. `V_(B)=9V,V_(C)=6V,V_(D)=11V`

Answer

B

Explanation

Potential at O is zero being earthed.
Applying kirchhoff's second law
`i(1+2+3)=12-6` or `i=1A`
`V_(A)-V_(D)=(1+2+3)xx1=6V`
`V_(A)-V_(B)=1xx1=1V`
`V_(A)-V_(C)=(1+2)xx1=3V`
Also, `V_(A)-V_(O)=12V` or `V_(A)=12V`
Thus, `V_(D)=12-6=6V`
`V_(B)=12-1=11V`, `V_(C)=12-3=9V`