Question

(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 \(12.5ms\)?

Short answer

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

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

Step-by-step solution

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.

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

Calculating the Time Constant

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 \(6.25\;ms\).

Calculating the Current

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 \(405\;mA\).