Bryson and Diego decide to work a practice problem.A resistor, inductor, and a battery are arranged in a circuit. The circuit has an inductance of L = 2 H, and a resistance of 2 kΩ. Switch S1 is suddenly closed at t = 0. Find the time needed for the current to reach a fraction f = 0.5 of its maximum value.
Question
Bryson and Diego decide to work a practice problem.A resistor, inductor, and a battery are arranged in a circuit. The circuit has an inductance of L = 2 H, and a resistance of 2 kΩ. Switch S1 is suddenly closed at t = 0. Find the time needed for the current to reach a fraction f = 0.5 of its maximum value.
Solution
The current in an LR circuit, when a battery is connected across it, grows according to the formula:
I(t) = I_max * (1 - e^(-t/τ))
where I_max is the maximum current, t is the time, and τ (tau) is the time constant of the circuit. The time constant τ is given by the ratio of the inductance L to the resistance R, i.e., τ = L/R.
Given that L = 2 H and R = 2 kΩ = 2000 Ω, we can calculate τ:
τ = L/R = 2 H / 2000 Ω = 0.001 s
We are asked to find the time t for the current to reach a fraction f = 0.5 of its maximum value. We can set up the equation:
0.5 = 1 - e^(-t/τ)
Solving for t gives:
ln(1 - 0.5) = -t/τ
t = -τ * ln(0.5) = 0.001 s * ln(0.5) ≈ 0.000693 s
So, it takes approximately 0.000693 seconds for the current to reach half of its maximum value.
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