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`        An inductor-coil of inductance 20 mH having resistance 10 Ω is joined to an ideal battery of emf 5.0 V. Find the rate of change of the induced emf at t= 0, (b) t = 10 ms and (c) t = 1.0 s.`
3 years ago

Deepak Patra
474 Points
```										Sol L = 20 mH; e =5.0 V, R = 10 Ω
Τ = L/R = 20x10^-3/10, i base 0 = 5/10
i = i0(1 – e^–t/τ)^2
⇒ i = i base 0 – i base 0 e^-t/τ^2
⇒ iR = i0R – i0Re^t / τ^2
(a) 10 x di/dt = d/dt i base 0 R + 10 x 5/10 x 10/ 20x 10^-3 x e^-0x10/2x10^-2
= 5/2 x 10^-3 x 1 = 5000/2 = 2500 = 2.5 x 10^-3 V/s.
b) Rdi/dt = R x i base  0 x 1/τ x e-t/τ
t = 10 ms = 10 x 10^-3 s
dE/dt = 10 x 5/10 x 10/20x 10^-3 x e^-0.01x 10/2x10^-2
= 16.844 = 17 V/’
c) For t = 1 s
dE/dt = Rdi/dt = 5/2 10^3 x e^10/2x10^-2 = 0.00 V/s.

```
3 years ago
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