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College Physics (Urone)
Found in: Page 863

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Short Answer

(a) What is the characteristic time constant of a \(25.0{\rm{ }}mH\) inductor that has a resistance of\(4.00{\rm{ }}\Omega \)? (b) If it is connected to a \(12.0{\rm{ }}V\) battery, what is the current after \(12.5ms\)?

a. What is the characteristic time constant is \(6.25\;ms\)

b. The current after \(12.5ms\) is \(405\;mA\).

See the step by step solution

Step by Step Solution

Step 1: Concept Introduction

Resistance is a measurement of the resistance to current flow in an electrical circuit. Resistance in ohms is denoted by the Greek letter omega \(\left( \Omega \right)\). Ohms are named after Georg Simon Ohm \(\left( {1784 - 1854} \right)\), a German scientist who studied the relationship between voltage, current, and resistance.

Step 2: Information Provided

  • Inductance of the magnet: \(25.0{\rm{ }}H\)
  • The resistance value: \(4.00{\rm{ }}\Omega \)
  • The battery voltage value: \(12.00{\rm{ }}V\)
  • The time value: \(12.5{\rm{ }}ms\)

Step 3: Calculating the Time Constant


The time constant will be given by

\(\tau = \frac{L}{R}\)

In our numerical case, we will have

\(\begin{array}{c}\tau = \frac{{0.025}}{4}\\ = 6.25\;ms\end{array}\)

Therefore, the required solution is \(6.25\;ms\).

Step 4: Calculating the Current


As can be seen, we must examine the intensity value after two-time constants. This value will be \({0.368^2} = 0.135\) twice the starting amount. Ohm's law may be used to determine the beginning value.

\(\begin{array}{c}I = \frac{U}{R}\\ = \frac{{12}}{4}\\ = 3\;A\end{array}\)

\(\begin{array}{c}I(12.5\;ms) = 0.135 \times 3\\ = 405\;mA\end{array}\)

Therefore, the current after the given time will be \(405\;mA\).

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