A 12.0-V battery is connected to a series circuit containing a 25.0 2 resistor and a 5.00 H inductor. In what time interval will the current reach 75.0%

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### Problem Statement

A 12.0-V battery is connected to a series circuit containing a 25.0 Ω resistor and a 5.00 H inductor. In what time interval will the current reach 75.0% of its maximum value?

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This problem involves an RL (resistor-inductor) circuit, which is a type of electrical circuit used to study the behavior of resistors and inductors when they are connected to a voltage source. The key aspect of this problem is to determine the time it takes for the current to reach 75% of its maximum value when a constant voltage is applied.

In an RL circuit, the growth of current \(I(t)\) over time \(t\) can be described using the following formula:

\[ 
I(t) = I_{max} \left(1 - e^{-t/\tau}\right) 
\]

where:
- \( I_{max} \) is the maximum current, which can be calculated as \( V/R \),
- \( \tau \) is the time constant, defined as \( L/R \),
- \( e \) is the base of the natural logarithm (approximately equal to 2.718).

Given:
- Battery voltage, \( V = 12.0 \, \text{V} \)
- Resistance, \( R = 25.0 \, \Omega \)
- Inductance, \( L = 5.00 \, \text{H} \)

Calculate \( I_{max} \):
\[ 
I_{max} = \frac{V}{R} = \frac{12.0 \, \text{V}}{25.0 \, \Omega} = 0.48 \, \text{A} 
\]

Calculate \( \tau \):
\[ 
\tau = \frac{L}{R} = \frac{5.00 \, \text{H}}{25.0 \, \Omega} = 0.20 \, \text{s} 
\]

To find the time \( t \) when the current reaches 75% of \( I_{max} \):
\[ 
0.75 \times I_{max} = I_{max} \left(1 - e^{-t/\tau}\right) 
\]

Solving for \( t \):
\[ 
0.75 = 1 - e^{-t/\tau} 
\]

\[
Transcribed Image Text:### Problem Statement A 12.0-V battery is connected to a series circuit containing a 25.0 Ω resistor and a 5.00 H inductor. In what time interval will the current reach 75.0% of its maximum value? --- This problem involves an RL (resistor-inductor) circuit, which is a type of electrical circuit used to study the behavior of resistors and inductors when they are connected to a voltage source. The key aspect of this problem is to determine the time it takes for the current to reach 75% of its maximum value when a constant voltage is applied. In an RL circuit, the growth of current \(I(t)\) over time \(t\) can be described using the following formula: \[ I(t) = I_{max} \left(1 - e^{-t/\tau}\right) \] where: - \( I_{max} \) is the maximum current, which can be calculated as \( V/R \), - \( \tau \) is the time constant, defined as \( L/R \), - \( e \) is the base of the natural logarithm (approximately equal to 2.718). Given: - Battery voltage, \( V = 12.0 \, \text{V} \) - Resistance, \( R = 25.0 \, \Omega \) - Inductance, \( L = 5.00 \, \text{H} \) Calculate \( I_{max} \): \[ I_{max} = \frac{V}{R} = \frac{12.0 \, \text{V}}{25.0 \, \Omega} = 0.48 \, \text{A} \] Calculate \( \tau \): \[ \tau = \frac{L}{R} = \frac{5.00 \, \text{H}}{25.0 \, \Omega} = 0.20 \, \text{s} \] To find the time \( t \) when the current reaches 75% of \( I_{max} \): \[ 0.75 \times I_{max} = I_{max} \left(1 - e^{-t/\tau}\right) \] Solving for \( t \): \[ 0.75 = 1 - e^{-t/\tau} \] \[
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