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Chapter 16: Problem 142

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 (A) \(0.1 \mathrm{~s}\) (B) \(0.05 \mathrm{~s}\) (C) \(0.3 \mathrm{~s}\) (D) \(0.15 \mathrm{~s}\)

Short Answer

Expert verified
The time for the current to reach half of its steady-state value in the given inductive circuit is approximately \(0.1 \mathrm{~s}\), which corresponds to answer (A).

Step by step solution

01

Write down the given values and formula for the current in an inductive circuit

We are given the inductance (L) as \(300 \mathrm{mH}\), the resistance (R) as \(2 \Omega\), and the voltage (V) as \(2 \mathrm{V}\). We will use the formula for the current in an inductive circuit, which is: \[i(t) = \frac{V}{R}(1 - e^{-\frac{tR}{L}})\] where i(t) is the current at time t, V is the voltage, R is the resistance, L is the inductance, and e is Euler's number (approximately 2.718).
02

Calculate the time constant

The time constant (蟿) for an inductive circuit is the ratio of the inductance (L) and the resistance (R): \[\tau = \frac{L}{R}\] We can plug in the given values and find the time constant: \[\tau = \frac{300 \times 10^{-3} \; \mathrm{H}}{2 \; \mathrm{\Omega}} = 0.15 \; \mathrm{s}\]
03

Set the current to half of its steady-state value and solve for the time

We know that the steady-state current value is \(\frac{V}{R}\). We want to find the time it takes for the current to reach half of this value: \[\frac{1}{2}\left(\frac{V}{R}\right) = \frac{V}{R}(1 - e^{-\frac{tR}{L}})\] Now, we can divide both sides by \(\frac{V}{R}\) and simplify: \[\frac{1}{2} = 1 - e^{-\frac{tR}{L}}\] We can now work to isolate the exponent: \[e^{-\frac{tR}{L}} = \frac{1}{2}\] We can then take the natural logarithm of both sides: \[-\frac{tR}{L} = \ln\left(\frac{1}{2}\right)\] Now, we can solve for t: \[t = -\frac{L \ln\left(\frac{1}{2}\right)}{R}\]
04

Plug in the values and find the time

Using the given values for L and R, we can calculate the time it takes for the current to reach half of its steady-state value: \[t = -\frac{300 \times 10^{-3} \; \mathrm{H} \times \ln\left(\frac{1}{2}\right)}{2 \; \mathrm{\Omega}}\] This simplifies to: \[t = \frac{0.3 \times \ln\left(\frac{1}{2}\right)}{-2} = 0.104 \; \mathrm{s}\] The time for the current to reach half of its steady-state value is approximately \(0.1 \mathrm{~s}\), which corresponds to answer (A).

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Most popular questions from this chapter

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