In the circuit diagram shown in Figure, `R=10 Omega, L=l5H, E=20v, i=2A`. This current is decreasing at a rate of `-1.0A//s.` Find `V_(ab)` at this instant.

Explanation

Potential difference across inductor is given as
`V_(L)=L(di)/(dt)=(5)(-1.0)=-5V`
Now, using Kirchhoff's second law, `V_(a)-iR-V_(L)-E=V_(b)`
`therefore" "V_(ab)=V_(a)-V_(b)=E+iR+V_(L)`
`=20+(2)(10)-5=35V`
Note In the above example, as the current is decreasing, the inductor can be replaced by a source of emf `e=|l.(di)/(dt)|`=5 V in such a manner that this emf supports the decreasing current, or it sends the current in the circuit in the same direction as the existing current. So, positive termianl of this source is towards b. Thus, the given circuit can be drawn as