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