The electric current i in a circuit containing a resistance and an inductance​ (see the figure​ below) is given by lnStartFraction i Over Upper I EndFraction equals StartFraction negative Rt Over Upper L EndFraction
​, where I is the current at tequals​0, R is the​ resistance, t is the​ time, and L is the inductance. Find how long​ (in s) it takes the current to reach 0.4 of the initial value of 0.759 ​A, if Iequals0.759 ​A, Requals7.56 Upper Omega​, and Lequals1.26 H.

We are given the equation for the electric current in a circuit containing resistance and inductance:

\[
\ln \left( \frac{i}{I} \right) = - \frac{R t}{L}
\]

where:
- \( i \) is the current at time \( t \),
- \( I \) is the current at \( t = 0 \),
- \( R \) is the resistance,
- \( L \) is the inductance,
- \( t \) is the time.

We are tasked to find the time \( t \) when the current reaches \( 0.4 \) of the initial current \( I \). That means:

\[
i = 0.4I
\]

### Step 1: Substitute the values
We can rewrite the equation using \( i = 0.4I \) and simplify:

\[
\ln \left( \frac{0.4I}{I} \right) = - \frac{Rt}{L}
\]

This simplifies to:

\[
\ln(0.4) = - \frac{Rt}{L}
\]

### Step 2: Isolate \( t \)

\[
t = - \frac{L \ln(0.4)}{R}
\]

Now, substitute the given values:
- \( R = 7.56 \, \Omega \),
- \( L = 1.26 \, \text{H} \),
- \( \ln(0.4) \approx -0.91629 \).

\[
t = - \frac{1.26 \times (-0.91629)}{7.56}
\]

Let me calculate this for you.The time it takes for the current to reach \(0.4\) of the initial value is approximately \(0.153\) seconds.

Would you like more details on the process, or do you have any further questions?

Here are 5 related questions for further exploration:
1. How would the time change if the resistance \(R\) were increased?
2. What happens to the current if \(L\) is doubled?
3. Can you calculate the current at a specific time \(t\), given these parameters?
4. How would this equation change for different inductive components?
5. What is the physical interpretation of the inductance \(L\) in the circuit?

**Tip**: The natural logarithm, \( \ln(x) \), is essential in solving decay processes, such as current in RL circuits!

The electric current i in a circuit containing a resistance and an inductance is given by ln(i/I) = -Rt/L, where I is the current at t=0, R is the resistance, t is the time, and L is the inductance. Find how long (in s) it takes the current to reach 0.4 of the initial value of 0.759 A, if I = 0.759 A, R = 7.56 Ω, and L = 1.26 H.

Learn how to calculate the time it takes for the current in an RL circuit to reach 0.4 of its initial value. This problem involves key concepts such as exponential decay, natural logarithms, and the equation ln(i/I) = -Rt/L. Given parameters include initial current (I = 0.759 A), resistance (R = 7.56 Ω), and inductance (L = 1.26 H).

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