## Related questions with answers

Model each application with a differential equation. All rates are assumed to be with respect to time. The voltage drop across an inductor is proportional to the rate at which the current is changing with respect to time.

Solution

VerifiedLet $V(t)$ be the voltage,and $I(t)$ the current on the inductor.,We know that the voltage can be expressed as

$\pmb{V(t)}=L\, \frac{dI(t)}{dt}$

where $L$ is the inductivity of the inductor.,The voltage change can thus be modeled as

$\frac{dV(t)}{dt}=L\, \frac{d^2I(t)}{dt^2}$

Since we have a voltage drop we will have a minus sign in front of $\frac{dV}{dt}$.,From the condition that the voltage drop across the inductor is proportional to the rate of change of the current with respect to time,we get the equation

$-\frac{dV(t)}{dt}=k\, \frac{dI(t)}{dt}\Leftrightarrow -L\frac{d^2I(t)}{dt^2}=k \, \frac{dI(t)}{dt}$

which can also be written as:

$\boxed{L\, \frac{d^2I(t)}{dt^2}+k\, \frac{dI(t)}{dt}=0}$

where $k$ is the proportionality constant.

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