For physics lab, two students constructed an RL circuit similar to the one shown in the figure, with = 6.00 V, L = 7.00 mH, and R = 5.80 Ω.A rectangular circuit contains a battery of emf ℰ on its left side, with the positive terminal above the negative terminal. An open switch S is on the top side, an inductor L is on the right side, and a resistor R is on the bottom side.(a)What is the inductive time constant of the circuit (in ms)? ms(b)Calculate the current in the circuit (in A) 250 µs after the switch is closed. A(c)What is the value of the final steady-state current (in A)? A(d)After what time interval (in ms) does the current reach 80.0% of its maximum value?
Question
For physics lab, two students constructed an RL circuit similar to the one shown in the figure, with = 6.00 V, L = 7.00 mH, and R = 5.80 Ω.A rectangular circuit contains a battery of emf ℰ on its left side, with the positive terminal above the negative terminal. An open switch S is on the top side, an inductor L is on the right side, and a resistor R is on the bottom side.(a)What is the inductive time constant of the circuit (in ms)? ms(b)Calculate the current in the circuit (in A) 250 µs after the switch is closed. A(c)What is the value of the final steady-state current (in A)? A(d)After what time interval (in ms) does the current reach 80.0% of its maximum value?
Solution
(a) The inductive time constant of the circuit (τ) is given by the formula τ = L/R. Substituting the given values, we get τ = 7.00 mH / 5.80 Ω = 1.21 ms.
(b) The current in the circuit after a time t after the switch is closed is given by the formula I = ε/R * (1 - e^(-t/τ)). Substituting the given values, we get I = 6.00 V / 5.80 Ω * (1 - e^(-250 µs / 1.21 ms)) = 0.85 A.
(c) The final steady-state current is given by the formula I = ε/R. Substituting the given values, we get I = 6.00 V / 5.80 Ω = 1.03 A.
(d) The time interval after which the current reaches 80.0% of its maximum value is given by the formula t = -τ * ln(1 - I/I_max). Substituting the given values, we get t = -1.21 ms * ln(1 - 0.80) = 1.77 ms.
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