The network shown in Fig. is a part of a complete circuit.What is the potential difference `V_(B)-V_(A)` when the current `I` is `5A` and is decreasing at a rate of `10^(2)(A//s)` ?

Answer

A

Explanation

`(di)/(dt)=10^(3) A//s`
`therefore` Induced `emf` across inductance
`|e|=(5xx10^(-3))(10^(3)) V`
`=5V`
Since, the current is decreasing the polarity of this `emf` would be so as to increase the existing current.The circuit can be redrawn as
Now `V_(A)-5+15+5=V_(B)`
`therefore V_(A)-V_(B)=-15V`
or `V_(B)-V_(A)=15 Vs`