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Differential equations play a crucial role in RL circuit analysis. They help us describe how the current changes over time in response to varying resistance and inductance. In our problem, the differential equation is set up as \[ L(t) \frac{di}{dt} + Ri = E(t) \] where

• \( L(t) \) is the time-varying inductance,
• \( R \) is resistance,
• \( E(t) \) is the external voltage.

What makes this equation special is how it incorporates the time dependence of the inductor. Solving these equations gives us detailed insights about current behavior under differing electrical components, each responding uniquely to initial conditions.

An inductor is a passive electrical component that stores energy in its magnetic field. Its behavior in a circuit is characterized by its inductance, which is the measure of how effectively it can store energy. For our RL circuit, the inductance is variable,\[ L(t) = 1 - \frac{1}{10}t \] for \( 0 \leq t < 10 \), and 0 thereafter. This introduces time-dependent dynamics to the circuit. The inductance decreasing over time means the inductor opposes changes in current less intense as time progresses. After \( t = 10 \), where inductance is zero, it no longer affects current, effectively removing itself from the circuit.

The current graph of an RL circuit visually represents how the current changes over time. Initially, for \( 0 \leq t < 10 \), the current increases from zero. This increase is governed by the differential equation:\[ \frac{di}{dt} = \frac{4 - 0.2i}{1 - \frac{1}{10}t} \]After solving, you'll notice the expected behavior of the current as it approaches a maximum or steady state. When \( t > 10 \), because the inductance is zero, the equation simplifies, and the current reaches and stays at a constant value. The graph typically would show a rise to this value and then level out, indicating how the inductance initially affects the current build-up.

A separable differential equation is a type of differential equation where variables can be separated on different sides of the equation. For the RL circuit, the differential equation given by:\[ \frac{di}{dt} = \frac{4 - 0.2i}{1 - \frac{1}{10}t} \] can be rearranged to\[ \frac{di}{4 - 0.2i} = \frac{dt}{1 - \frac{1}{10}t} \] Here, each side of the equation depends on a single variable, either \( i \) or \( t \). Separating and integrating each side allows us to solve for \( i(t) \) given the initial condition \( i(0) = 0 \). Solving such equations is essential for understanding how RL circuits transition current over time, combining mathematical techniques with physical electrical concepts.