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Expert-verified Found in: Page 863 ### College Physics (Urone)

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 $$405\;mA$$.

See the 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

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 $$6.25\;ms$$.

## Step 4: 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 $$405\;mA$$. ### Want to see more solutions like these? 