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In the circuit diagram shown in Figure, ...

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.

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`PD` across inductor,
`V_L= L (di)/(dt)=(5)(-1.0)=-5V`
now, `V_a-iR-V_L=E=V_b`
`:. V_(ab) =V_a-V_b=E+iR+V_L`
`=20+(2)(10)-5=35V`
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