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Chapter 14: Problem 2051

A coil of inductance \(300 \mathrm{mH}\) and resistance \(2 \Omega\) is connected to a source of voltage \(2 \mathrm{~V}\). The current reaches half of its steady state value in \(\ldots \ldots\) (a) \(0.15 \mathrm{sec}\) (b) \(0.3 \mathrm{sec}\) (c) \(0.05 \mathrm{sec}\) (d) \(0.1 \mathrm{sec}\)

Short Answer

Expert verified
(d) \(0.1 \mathrm{sec}\)

Step by step solution

01

Calculate the Time Constant 蟿

To calculate the time constant 蟿 (tau) of the RL circuit, we will use the formula 蟿 = L/R, where L is the inductance (300 mH) and R is the resistance (2 惟). In order to calculate the time constant in seconds, we need to convert the inductance to henries (H). \( L = 300 mH 脳 \frac{1 H}{1000 mH} = 0.3 H \) Now, let's calculate the time constant 蟿: \( 蟿 = \frac{L}{R} = \frac{0.3 H}{2 惟} = 0.15 s \)
02

Calculate the Steady-State Current

Next, we need to find the steady-state current using Ohm's law (I = V/R), where V is the voltage (2 V) and R is the resistance (2 惟). \( I_{steady-state} = \frac{V}{R} = \frac{2 V}{2 惟} = 1 A \)
03

Find the Time it Takes to Reach Half of the Steady-State Current

Now, our goal is to find the time it takes for the current to reach half of the steady-state value: \( 0.5 脳 I_{steady-state} = I_{steady-state} 脳 (1 - e^{\frac{-t}{蟿}}) \) Dividing both sides by the steady-state current: \( 0.5 = 1 - e^{\frac{-t}{蟿}} \) Now solve for t: \( 0.5 = e^{\frac{-t}{蟿}} \) Applying the natural logarithm to both sides: \( -ln(2) = \frac{-t}{蟿} \) Multiplying both sides by 蟿: \( -蟿 ln(2) = t \) Now we can substitute 蟿 (0.15 s) to find the time (t): \( t = -(0.15 s) 脳 ln(2) 鈮� 0.104 s \) Since the time is closest to 0.1 s, the correct answer is: (d) \(0.1 \mathrm{sec}\)

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