Question

In the circuit in the figure, \(R=120 . \Omega, L=3.00 \mathrm{H},\) and \(V_{\mathrm{emf}}=40.0 \mathrm{~V}\) After the switch is closed, how long will it take the current in the inductor to reach \(300 . \mathrm{mA} ?\)

Short answer

Answer: Approximately 0.0692 seconds.

Step-by-step solution

Identify the known variables

In this circuit, we are given the following known variables: R = 120 Ω (resistor) L = 3.00 H (inductance) V_emf = 40.0 V (voltage) I = 300 mA = 0.3 A (target current)

Write down the formula for current in an RL circuit

The time-dependent current in an RL circuit after the switch is closed can be expressed as: $$ I(t) = \frac{V_{emf}}{R}(1-e^{-\frac{Rt}{L}}) $$

Substitute the known values and solve for time t

Now we can substitute the known values and the target current of 0.3 A into the equation to solve for time t: $$ 0.3 = \frac{40}{120}(1-e^{-\frac{120t}{3}}) $$

Simplify and isolate the exponential term

To isolate the exponential term, we first divide both sides by 40/120: $$ \frac{0.3}{\frac{40}{120}} = 1-e^{-\frac{120t}{3}} $$ Simplify the equation: $$ \frac{9}{10} = 1 - e^{-\frac{120t}{3}} $$ Subtract 9/10 from both sides to isolate the exponential term: $$ e^{-\frac{120t}{3}} =1-\frac{9}{10} = \frac{1}{10} $$

Take the natural logarithm of both sides

Now we can take the natural logarithm of both sides to get rid of the exponential term: $$ -\frac{120t}{3}=\ln{\frac{1}{10}} $$

Solve for t

Now we can solve for t by multiplying both sides by -3/120: $$ t = -\frac{3}{120} \ln{\frac{1}{10}} $$ Calculate the result: $$ t \approx 0.0692 \text{ seconds} $$ So it takes approximately 0.0692 seconds for the current to reach 300 mA after closing the switch.