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I = $\frac{V}{R}$(1 - $e^{-(R/L)t}$)

Please check the explanation for derivation

Explanation:

The solution:

Given: Voltage drop across resistor + inductor = total EMF

$⟹$ RI+L$\frac{dI}{dt}$ ​=V

L$\frac{dI}{dt}$ ​=V - RI

$\frac{dI}{dt}$ = $\frac{V-RI}{L}$

Multiply both sides by dt and divide both by (V - RI):

$\frac{dI}{V-RI}$ = $\frac{dt}{L}$

Integrating both sides:

$\int$$\frac{dI}{V-RI}$ = ∫$\frac{dt}{L}$

$⟹$ $-\frac{ln(V-RI)}{R}$ = $\frac{1}{L}t$ + K

Given initial condition I(0) = 0

We get, K = $-\frac{lnV}{R}$

Substituting K back into our expression:

$⟹$ $-\frac{ln(V-RI)}{R}$ = $\frac{1}{L}t$ $-\frac{lnV}{R}$

Rearranging:

$\frac{lnV}{R}$ - $\frac{(V-RI)}{R}$ = $\frac{1}{L}t$

Multiplying throughout by -R:

−ln V+ln(V−RI)= $-\frac{R}{L}t$

Collecting the logarithm parts together:

$ln(\frac{V-RI}{V})$ = $-\frac{R}{L}t$

Taking "e to both sides":

$\frac{V-RI}{V}$ = $e^{-(R/L)t}$

1 - $\frac{R}{V}I$ = $e^{-(R/L)t}$

- $\frac{R}{V}I$ = -1 + $e^{-(R/L)t}$

I = $\frac{V}{R}$(1 - $e^{-(R/L)t}$)

Q factor can be found only when the frequency of alternating power source is provided.

Q = P_stored/P_dissipated = I²X/I²R

Q = X/R

Where X is the inductive reactance