When a direct voltage is connected to a circuit, the current does not start flowing smoothly immediately, but follows the equation
L*(dI/dt)+RI=V
where L is the inductance of the circuit, R is the resistance, I is the current and V is the voltage. Voltage
is switched at time t = 0, so that I(0) = 0. Determine I(t) and its limit value
limt→∞ I(t). When has the current in the circuit reached 50% of this limit value?

To solve the equation given by 

\[
L \frac{dI}{dt} + RI = V,
\]

we'll follow these steps:

1. **Rewrite the equation**: This is a first-order linear ordinary differential equation (ODE). Rearranging it gives:

\[
\frac{dI}{dt} + \frac{R}{L}I = \frac{V}{L}.
\]

2. **Identify the integrating factor**: The integrating factor, \( \mu(t) \), is given by:

\[
\mu(t) = e^{\int \frac{R}{L} dt} = e^{\frac{R}{L}t}.
\]

3. **Multiply the ODE by the integrating factor**:

\[
e^{\frac{R}{L}t} \frac{dI}{dt} + \frac{R}{L} e^{\frac{R}{L}t} I = \frac{V}{L} e^{\frac{R}{L}t}.
\]

4. **Rewrite the left-hand side**:

\[
\frac{d}{dt} \left( e^{\frac{R}{L}t} I \right) = \frac{V}{L} e^{\frac{R}{L}t}.
\]

5. **Integrate both sides**:

\[
e^{\frac{R}{L}t} I = \int \frac{V}{L} e^{\frac{R}{L}t} dt = \frac{V}{R} e^{\frac{R}{L}t} + C,
\]

where \( C \) is a constant of integration.

6. **Solve for \( I(t) \)**:

\[
I(t) = \frac{V}{R} + Ce^{-\frac{R}{L}t}.
\]

7. **Apply the initial condition \( I(0) = 0 \)**:

At \( t = 0 \):

\[
0 = \frac{V}{R} + C \implies C = -\frac{V}{R}.
\]

Thus, the expression for \( I(t) \) becomes:

\[
I(t) = \frac{V}{R} - \frac{V}{R} e^{-\frac{R}{L}t} = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right).
\]

### Limit of \( I(t) \)

To find the limit as \( t \) approaches infinity:

\[
\lim_{t \to \infty} I(t) = \frac{V}{R} \left( 1 - 0 \right) = \frac{V}{R}.
\]

### Time to reach 50% of the limit value

To find when \( I(t) \) reaches 50% of its limit value \( \frac{V}{R} \):

\[
0.5 \cdot \frac{V}{R} = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right).
\]

Dividing both sides by \( \frac{V}{R} \) (assuming \( V \neq 0 \)) gives:

\[
0.5 = 1 - e^{-\frac{R}{L}t}.
\]

Solving for \( e^{-\frac{R}{L}t} \):

\[
e^{-\frac{R}{L}t} = 0.5 \implies -\frac{R}{L}t = \ln(0.5) \implies t = -\frac{L}{R} \ln(0.5).
\]

This simplifies to:

\[
t = \frac{L}{R} \ln(2).
\]

### Summary

- The current \( I(t) \) is given by:

\[
I(t) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L}t} \right).
\]

- The limit value is \( \frac{V}{R} \).
- The time to reach 50% of this limit is \( t = \frac{L}{R} \ln(2) \).

Do you want more details or have any questions? Here are some related questions you might consider:

1. What physical principles are illustrated by the behavior of current in this circuit?
2. How does changing the resistance \( R \) affect the time to reach 50% of the limit value?
3. What happens to the current if the voltage \( V \) is increased?
4. How does the inductance \( L \) influence the transient response of the circuit?
5. Can you explain the concept of time constant in the context of this circuit?

**Tip**: Understanding how inductors and resistors interact in circuits can provide insights into more complex electrical systems.

When a direct voltage is connected to a circuit, the current does not start flowing smoothly immediately, but follows the equation L*(dI/dt)+RI=V. Determine I(t) and its limit value limt→∞ I(t). When has the current in the circuit reached 50% of this limit value?

Explore the dynamics of current flow in an RL circuit governed by the equation L*(dI/dt)+RI=V. This detailed analysis includes determining the function I(t), its limit as time approaches infinity, and the time taken to reach 50% of the limit value.

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