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Chapter 32: Q. 7 (page 924)

The peak current to and from a capacitor is $10 m A$ . What is the peak current if
a. The emf frequency is doubled?
b. The emf peak voltage is doubled (at the original frequency)

Short Answer

Expert verified

Result is:

$a.) I_{o}^{'} = 20 mA$

$b.) I_{o}^{'} = 20 mA$

Step by step solution

01

Part (a) Step 1: Given information

We have been given that $I_{o} = 10 mA$

we will use the formula to sovle is:

$I_{o} = (2 蟺 f) V_{o} C and I_{o} 伪 f$

02

Part(a) Step 2: Simplify

if $f$ is the frequency doubled:

$I_{o}^{'} = 2 路 I_{o}$

As, $I_{o}^{} = 10 m A$

$I_{o}^{'} = 2 路 10 m A$

$I_{o}^{'} = 20 m A$

03

Part (b) Step 1: Given information

We have been given that $I_{0} = 10 mA$

According to the formula:

$I_{o} = (2 蟺 f) V_{o} C and I_{o} 伪 f$

04

Part (b) Step 2: Simplify

$I_{o} 伪 V_{o} 鈬� V_{o} (doubled) 鈬� I_{o}^{'} = 2 I_{o}$

$I_{o}^{'} = 2 路 10 mA$

$I_{o}^{'} = 20 m A$

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Most popular questions from this chapter

Lightbulbs labeled $40 W, 60 W$ and $100 W$ are connected to a $120 V / 60 H z$ power line as shown in the figure .What is the rate at which energy is dissipated in each bulb?

The telecommunication circuit shown in FIGURE CP $32.71$ has a parallel inductor and capacitor in series with a resistor.

a. Use a phasor diagram to show that the peak current through the resistor is

$I = \frac{蔚_{0}}{\sqrt{R^{2} + {(\frac{1}{X_{L}} - \frac{1}{X_{C}})}^{- 2}}}$

b. What is I in the limits $蝇鈫� 0$ and $蝇鈫� 鈭�$ ?

c. What is the resonance frequency $蝇_{0}$ ? What is I at this frequency?

FIGURE $Q 32.1$ shows emf phasors a, b, and c.

a. For each, what is the instantaneous value of the emf?

b. At this instant, is the magnitude of each emf increasing, decreasing, or holding constant?

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

A series RLC circuit attached to a $120 V / 60 H z$ power line draws a $2.4 A$ rms current with a power factor of $0.87$ . What is the value of the resistor?

See all solutions

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