Chapter 30: Q. 67 (page 874)

67. II FIGURE P30.67 shows the potential difference across a potential difference across a $50 mH$ inductor. The current through the inductor at $t = 0 s$ is $0.20 A$ . Draw a graph showing the current through the inductor from $t = 0 s$ to $t = 40 ms$

Short Answer

Expert verified

The following graph between current and time is

Step by step solution

Step 1. Given information

We have $Δ r_{L} = - 1 V_{for} 0 ms < t < 10 ms$

$\begin{array}{l} So \\ - L \frac{d I}{d t} = - 1 V \\ \frac{d I}{d t} = \frac{1 V}{50 mH} = 20 A / s \end{array}$

$\begin{array}{l} d I = (20 A / s) d t I = (20 A / s) t + c \\ Butat t = 0, I = 0.2 A \\ 0.2 A = 0 + c \end{array}$

$\begin{array}{l} I = (20 A / s) t + 0.2 A \\ At t = 10 ms = 10 \times 10^{- 3} s \\ I = (20 A / s) (10 \times 10^{- 3} s) + 0.2 A \\ = 0.2 A + 0.2 A \\ = 0.4 A \end{array}$

Explanation

From $t = 0 ms$ to $t = 10 ms$ current increase from $0.2 A$ to $0.4 A$ .

From the figure below, for $Δ V_{L} = 0$

for $10 ms < t < 20 ms$ $I = 0.4 A$ $10 ms < t < 20 ms$

Now $Δ V_{L} = 2 V_{for} 20 ms < t < 30 ms$

$\begin{array}{l} - L \frac{d I}{d t} = 2 V \\ \frac{d I}{d t} = \frac{- 2 V}{50 mH} = - 40 A / s (or) \\ d I = (- 40 A / s) d t \\ I = (- 40 A / s) t + c \\ At t = 20 ms, \\ I = 0.4 A \\ 0.4 A = (- 40 A / s) (20 ms) (10^{- 3} s / ms) + c \\ \Rightarrow 0.4 A = - 0.8 A + c \\ \Rightarrow c = 1.2 A I \\ = (- 40 A / s) t + 1.2 A \end{array}$

Now At

$t = 30 ms = 30 \times 10^{- 3} s$

$\begin{array}{l} I = (- 40 A / s) (30 \times 10^{- 3} s) + 1.2 A \\ = - 1.2 A + 1.2 A \\ = 0 A \end{array}$

Current decreases from $0.4 A_{to} 0 A_{in}$ the time $20 ms < t < 30 ms$

After $t = 30 ms$ current remains at $0 A$

So we can have the following graph:

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67. II FIGURE P30.67 shows the potential difference across a potential difference across a $50 mH$ inductor. The current through the inductor at $t = 0 s$ is $0.20 A$ . Draw a graph showing the current through the inductor from $t = 0 s$ to $t = 40 ms$

Short Answer

Step by step solution

Step 1. Given information

Explanation

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67. II FIGURE P30.67 shows the potential difference across a potential difference across a 50mHinductor. The current through the inductor at t=0sis 0.20A. Draw a graph showing the current through the inductor from t=0sto t=40ms

Short Answer

Step by step solution

Step 1. Given information

Explanation

One App. One Place for Learning.

Most popular questions from this chapter

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Physics of Motion

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Study anywhere. Anytime. Across all devices.

67. II FIGURE P30.67 shows the potential difference across a potential difference across a $50 mH$ inductor. The current through the inductor at $t = 0 s$ is $0.20 A$ . Draw a graph showing the current through the inductor from $t = 0 s$ to $t = 40 ms$