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Q74PE

Expert-verifiedFound in: Page 863

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(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 \(405\;mA\).

**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.**

- 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: \(12.5{\rm{ }}ms\)

a)

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 \(6.25\;ms\).

b)

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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