Q. 66 66. II FIGURE P30.66 shows the c... [FREE SOLUTION]

Chapter 30: Q. 66 (page 873)

66. II FIGURE P30.66 shows the current through a $10 mH$ inductor. Draw a graph showing the potential difference $螖 V_{L}$ across the inductor for these $6 ms$ .

Short Answer

Expert verified

The potential difference across the inductor is

$\begin{array}{l} 螖 V_{L} = 鈭� 10 Vfor 0 ms < t < 2 ms \\ 螖 V_{L} = 0 痴鈥勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌刦辞谤 2 ms < t < 3 ms \\ 螖 V_{L} = 20 痴鈥勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌刦辞谤 3 ms < t < 5 ms \\ 螖 V_{L} = 0 痴鈥勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌刦辞谤 t > 5 ms \end{array}$

Step by step solution

Step 1. Introduction

It is given that the value of inductor, $L = 10 mH = 10 脳 10^{鈭� 3} H$

We have potential difference across the inductor as $螖 V_{L} = 鈭� L \frac{d I}{d t}$

Step 2. Explanation

From $t = 0 ms$ to $t = 2 ms$ , current $I$ changes from $0 鈥凙鈥则辞鈥�2础$ .

$\begin{array}{l} \frac{d I}{d t} = \frac{(2 A 鈭� 0 A)}{(2 ms 鈭� 0 ms)} = \frac{2 A}{(2 ms) (10^{鈭� 3} s / ms)} = 1000 A / s \\ 螖 V_{L} = 鈭� (10 脳 10^{鈭� 3} H) (1000 A / s) = 鈭� 10 V \end{array}$

Step 3.

From $t = 2 ms$ to role="math" $t = 3 ms,$ role="math" $I$ remains constant

Therefore voltage becomes zero.

Step 3.

Now from $t = 3 ms$ to $5 ms$ , the current decreases from $2 A$ to $鈭� 2 A$

$\begin{array}{l} \frac{d I}{d t} = \frac{(鈭� 2 A 鈭� 2 A)}{(5 ms 鈭� 3 ms)} \\ = \frac{鈭� 4 A}{(2 ms) (10^{鈭� 3} s / ms)} \\ = 鈭� 2000 A / s \end{array}$