We are given an RL circuit with the following parameters:

• Voltage (emf): $V(t) = 3 \sin(2t)$
• Resistance $R = 10 \, \Omega$
• Inductance $L = 0.5 \, H$
• Initial current $I(0) = 6 \, \text{A}$

The equation that governs the RL circuit is given by the first-order linear differential equation derived from Kirchhoff's voltage law:

$L \frac{dI}{dt} + RI = V(t)$

Substituting the given values:

$0.5 \frac{dI}{dt} + 10I = 3 \sin(2t)$

Step 1: Solve the homogeneous equation

First, solve the homogeneous differential equation:

$0.5 \frac{dI}{dt} + 10I = 0$

This simplifies to:

$\frac{dI}{dt} + 20I = 0$

This is a separable differential equation. Solving it:

$\frac{dI}{I} = -20 dt$

Integrating both sides:

$\ln |I| = -20t + C_1$

Exponentiating both sides:

$I_h(t) = C_2 e^{-20t}$

where $C_2 = e^{C_1}$ is a constant to be determined later using the initial condition.

Step 2: Solve the particular solution

We now find a particular solution to the non-homogeneous equation. We try a solution of the form:

$I_p(t) = A \cos(2t) + B \sin(2t)$

Substitute $I_p(t)$ into the original non-homogeneous equation:

$0.5 \left( -2A \sin(2t) + 2B \cos(2t) \right) + 10 \left( A \cos(2t) + B \sin(2t) \right) = 3 \sin(2t)$

Simplifying:

$- A \sin(2t) + B \cos(2t) + 10A \cos(2t) + 10B \sin(2t) = 3 \sin(2t)$

Group the terms involving $\sin(2t)$ and $\cos(2t)$:

$(-A + 10B) \sin(2t) + (B + 10A) \cos(2t) = 3 \sin(2t)$

Equating coefficients of $\sin(2t)$ and $\cos(2t)$:

1. $-A + 10B = 3$
2. $B + 10A = 0$

Step 3: Solve for constants $A$ and $B$

From the second equation, $B = -10A$. Substituting this into the first equation:

$-A + 10(-10A) = 3$

$-A - 100A = 3$

$-101A = 3 \quad \Rightarrow \quad A = -\frac{3}{101}$

Now, substitute $A = -\frac{3}{101}$ into $B = -10A$:

$B = -10 \left( -\frac{3}{101} \right) = \frac{30}{101}$

Thus, the particular solution is:

$I_p(t) = -\frac{3}{101} \cos(2t) + \frac{30}{101} \sin(2t)$

Step 4: General solution

The general solution is the sum of the homogeneous and particular solutions:

$I(t) = I_h(t) + I_p(t) = C_2 e^{-20t} - \frac{3}{101} \cos(2t) + \frac{30}{101} \sin(2t)$

Step 5: Apply the initial condition

We are given $I(0) = 6$. Substituting $t = 0$ into the general solution:

$6 = C_2 e^{0} - \frac{3}{101} \cos(0) + \frac{30}{101} \sin(0)$

$6 = C_2 - \frac{3}{101}$

Solving for $C_2$:

$C_2 = 6 + \frac{3}{101} = \frac{606 + 3}{101} = \frac{609}{101}$

Final Solution

The current in the circuit at any time $t$ is:

$I(t) = \frac{609}{101} e^{-20t} - \frac{3}{101} \cos(2t) + \frac{30}{101} \sin(2t)$

This is the complete solution for the current as a function of time.

Would you like any further details or explanations?

Here are 5 related questions:

1. How does the inductance affect the behavior of an RL circuit?
2. What would happen to the solution if the resistance was increased?
3. How does the particular solution represent the steady-state response of the circuit?
4. Can you explain the difference between transient and steady-state behavior in this context?
5. What role does the initial condition play in determining the total solution?

Tip: In RL circuits, the transient response is often dominated by the inductor, while the steady-state response is determined by the resistor.