A 40-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 60 ohms. Find the current 
i(t)
 if 
i(0) = 0.

i(t) = 
 


Determine the current as 
t → ∞.

lim 
t→∞ 
i(t) = 
 


The given problem describes an LR-series circuit where the electromotive force (EMF) is 40 volts, the inductance \( L = 0.1 \) henry, the resistance \( R = 60 \) ohms, and the initial current \( i(0) = 0 \). The goal is to find the current \( i(t) \) as a function of time and determine its behavior as \( t \to \infty \).

### Step 1: Write the Differential Equation

The current in an LR circuit follows the first-order linear differential equation:

\[
L \frac{di(t)}{dt} + R i(t) = V(t)
\]

where:
- \( L = 0.1 \) henry (inductance),
- \( R = 60 \) ohms (resistance),
- \( V(t) = 40 \) volts (constant electromotive force).

Substituting the given values into the equation:

\[
0.1 \frac{di(t)}{dt} + 60 i(t) = 40
\]

### Step 2: Solve the Differential Equation

First, divide through by \( 0.1 \) to simplify:

\[
\frac{di(t)}{dt} + 600 i(t) = 400
\]

This is a linear differential equation, and we can solve it using an integrating factor.

The integrating factor is:

\[
\mu(t) = e^{\int 600 \, dt} = e^{600t}
\]

Multiply both sides of the differential equation by \( e^{600t} \):

\[
e^{600t} \frac{di(t)}{dt} + 600 e^{600t} i(t) = 400 e^{600t}
\]

The left-hand side is the derivative of \( e^{600t} i(t) \):

\[
\frac{d}{dt} \left( e^{600t} i(t) \right) = 400 e^{600t}
\]

Integrate both sides with respect to \( t \):

\[
e^{600t} i(t) = \frac{400}{600} e^{600t} + C
\]

Simplify:

\[
e^{600t} i(t) = \frac{2}{3} e^{600t} + C
\]

Now, divide both sides by \( e^{600t} \):

\[
i(t) = \frac{2}{3} + C e^{-600t}
\]

### Step 3: Apply Initial Condition

We are given that \( i(0) = 0 \). Using this:

\[
0 = \frac{2}{3} + C e^{0}
\]

\[
0 = \frac{2}{3} + C
\]

\[
C = -\frac{2}{3}
\]

Thus, the solution for \( i(t) \) is:

\[
i(t) = \frac{2}{3} - \frac{2}{3} e^{-600t}
\]

### Step 4: Find the Limit as \( t \to \infty \)

As \( t \to \infty \), the exponential term \( e^{-600t} \to 0 \). Therefore, the current \( i(t) \) approaches:

\[
\lim_{t \to \infty} i(t) = \frac{2}{3} \, \text{amperes}.
\]

### Final Answer:

- The current as a function of time is:

\[
i(t) = \frac{2}{3} \left( 1 - e^{-600t} \right)
\]

- As \( t \to \infty \), the current approaches:

\[
\lim_{t \to \infty} i(t) = \frac{2}{3} \, \text{amperes}.
\]

---

Would you like further details on any step, or do you have any other questions?

Here are 5 related questions you might find interesting:
1. How does the time constant in an LR circuit affect the current response?
2. What happens if the resistance in the circuit increases?
3. How does inductance impact the time for the current to reach its steady-state value?
4. Can the solution be extended for a time-varying voltage source \( V(t) \)?
5. How does the energy stored in the inductor change over time?

**Tip**: The time constant \( \tau = \frac{L}{R} \) controls how fast the current approaches its steady-state value in an LR circuit.

A 40-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 60 ohms. Find the current i(t) if i(0) = 0. Determine the current as t → ∞.

Find the current i(t) in an LR-series circuit with a 40-volt electromotive force, 0.1 henry inductance, and 60 ohms resistance. Learn how to solve the differential equation for the current as a function of time and determine the current as t approaches infinity. This problem covers LR circuits and exponential decay, suitable for first-year undergraduate students.

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