1.

All resistance shown in the circuit are `2Omega` each. The current in the resistance between `D` and `E` is A. `5A`B. `2.5A`C. `1A`D. `7.5A`

Answer» Correct Answer - B
Resistance between `A` and `B` can be removed due to balanced wheastone bridfge concept. Now, `R_(DE)` and `R_(GH)` are in series and they are connected in parallel with `10 V` battery.
`:. I_(DE)=10/(R_(DE)+R_(HG))=10/(2+2)`
`=2.5A`


Discussion

No Comment Found

Related InterviewSolutions