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The network shown in the figure is a part of complete circuit. What is the potential difference `V_B-V_A` when the current `I` is `5 A` and is decreasing at a rate of `10^3A//s`?

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The correct Answer is:
A
`(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`

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