Question

Consider an \(\mathrm{RL}\) circuit with resistance \(R=1.00 \mathrm{M} \Omega\) and inductance \(L=1.00 \mathrm{H},\) which is powered by a \(10.0-\mathrm{V}\) battery. a) What is the time constant of the circuit? b) If the switch is closed at time \(t=0\), what is the current just after that time? After \(2.00 \mu\) s? When a long time has passed?

Short answer

a) The time constant of the RL circuit is 1.00 μs. b) The current in the circuit: - Immediately after the switch is closed is 0 A. - After 2.00 μs is approximately 6.32 mA. - When a long time has passed is 10.0 mA.

Step-by-step solution

Identify the given information.

We have the resistance R = 1.00 MΩ, inductance L = 1.00 H, and the voltage across the battery V = 10.0 V.

Calculate the time constant.

The time constant is given by the formula \(\tau = \frac{L}{R}\). Plug in the given values: \(\tau = \frac{1.00 \ \text{H}}{1.00 \ \text{M} \Omega} = \frac{1.00 \ \text{H}}{1.00 \times 10^6 \ \Omega} = 1.00 \times 10^{-6} \ \text{s} = 1.00 \ \mu \text{s}\). So, the time constant \(\tau\) is 1.00 μs.

Calculate the current immediately after the switch is closed.

The current in the RL circuit as a function of time is given by: \(I(t) = \frac{V}{R} (1 - e^{-t/\tau})\) At \(t=0\), the current is \(I(0) = \frac{10.0 \ \text{V}}{1.00 \ \text{M} \Omega} (1 - e^{-0/(1.00 \ \mu \text{s})} ) = 0 \ \text{A}\). The current just after closing the switch is 0 A.

Calculate the current after 2.00 μs.

Using the current equation, replace t with 2.00 μs: \(I(2.00 \ \mu \text{s}) = \frac{10.0 \ \text{V}}{1.00 \ \text{M} \Omega} (1 - e^{-2.00 \ \mu \text{s}/(1.00 \ \mu \text{s})}) \approx 6.32 \ \text{mA}\) The current after 2.00 μs is approximately 6.32 mA.

Calculate the current when a long time has passed.

When a long time has passed, the exponential factor in the current equation goes to zero. Hence, \(I(\infty) = \frac{V}{R} (1 - e^{-\infty}) = \frac{10.0 \ \text{V}}{1.00 \ \text{M} \Omega} = 10.0 \ \text{mA}\) The current when a long time has passed is 10.0 mA.