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Chapter 32: Question 25 (page 925)

An inductor is connected to a $15 K H z$ oscillator. The peak current is $65 m A$ when the rms voltage is width="36">6.0V. What is the value of the inductance L?

Short Answer

Expert verified

The value of inductance $(L) = 1.37 脳 10^{- 3} 鈮� 1.4 脳 10^{- 3}$

Step by step solution

01

GIVEN INFORMATION:

Frequency of oscillator $(f) = 15 K H z$

peak current $(I_{慰}) = 65 m A$

RMS voltage $(V_{r m s}) = 6.0 V$

Here we need to find the inductance of the inductor through the formula $X_{L} = 2 蟺 f L$ where $X_{L}$ is the inductive reactance of the inductor.

02

EXPLANATION:

We know that $I_{rms} = \frac{I_{慰}}{\sqrt{2}}$

and $X_{L} = \frac{V_{r m s}}{I_{rms}}$

$= \frac{6 \sqrt{2}}{65 脳 10^{- 3}}$ (substituting values)

$= 130 惟$

now, $X_{L} = 2 蟺 f L$

$L = \frac{X_{L}}{2 蟺 f}$

$= \frac{130}{2 蟺脳 15000}$ (Substituting values)

$= 1.37 脳 10^{- 3} H$

$鈮� 1.4 脳 10^{- 3} H$

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Most popular questions from this chapter

A series RLC circuit consists of a $50 惟$ resistor, a localid="1650491262241" $3.3 m H$ inductor, and a $480 n F$ capacitor. It is connected to a localid="1650491276532" $5.0 k H z$ oscillator with a peak voltage of localid="1650491288645" $5.0 V$ . What is the instantaneous current $i$ when

a. $蔚 = 蔚_{0}$ ?

b. $蔚 = 0 V$ and is decreasing?

For an RC filter circuit, show that $V_{R} = V_{C} = \frac{蔚_{0}}{\sqrt{2}}$ at $蝇 = 蝇_{c}$ .

A high-pass RC filter is connected to an AC source with a peak voltage of $10.0 V$ . The peak capacitor voltage is $6.0 V$ . What is the resistor voltage?

A $20 n F$ capacitor is connected across an AC generator that produces a peak voltage of $5.0 V$ .

a. At what frequency $f$ is the peak current $50 m A$ ?

b. What is the instantaneous value of the emf at the instant when $i_{c} = I_{c}$ ?

In FIGURE, what is the current supplied by the emf when

(a) the frequency is very small and

(b) the frequency is very large

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