Ask question

Ask question

All categories

1 year ago

12

The unknown quantities to be determined are (a) the capacitive reactance, (b) the maximum and the rms voltages from the source,

and (c) the rms current in the capacitor.
Evaluate the capacitive or inductive reactance, XC or XL.

The unknown quantity to be determined in part (a) is the circuit's capacitive reactance
XC defined as
XC ≡ 1/2fC
,
where, in this problem, C = 4.48 F and the AC source frequency f must be determined.

What is the frequency f of the AC source?

1 answer:

Digiron [165]1 year ago

8 0

The frequency f of the AC source is determined as 0.446 Xc.

<h3>Frequency of the AC source</h3>

The frequency of the AC source is calculated as follows;

Xc ≡ 1/2fC

where;

Xc is the capacitive reactance
f is frequency
C is capacitance

fC = 2Xc

f = 2Xc / C

f = (2Xc)/4.48

f = 0.446 Xc

Thus, the frequency f of the AC source is determined as 0.446 Xc.

Learn more about frequency here: brainly.com/question/10728818

#SPJ1

You might be interested in

If the speed of the magnet is doubled, the induced voltage is ________ . twice as great four times as great half as great unchan

Rina8888 [55]

If the speed of the magnet is doubled, the induced voltage is twice as great

Faraday's law of induction states that whenever relative motion exists between a coil and magnetic field, a voltage will induce in the circuit and this voltage is proportional to the rate of change of the flux. The expression for the motional emf is as follows:

ε=Blv

Here, ε is the motional emf, B is the magnetic field, l is the length of the conductor, and v is the velocity at which the magnetic field changes.

The induced voltage by the moving magnet is directly proportional to the speed of the magnet. Therefore, an increase in the speed of the magnet will increase the induced voltage.

If the speed of the magnet is doubled, the induced voltage is twice as great

Find out more at: brainly.com/question/13369951

4 0

1 year ago

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a

Nat2105 [25]

Answer:

Explanation:

Given

diameter $d=20 \mu m$

density $\rho =1300 kg/m^3$

frequency $\nu =150 Hz$

Length of silk strand $L=14 cm$

Velocity in the string is as follows

$\nu =\sqrt{\frac{T}{\mu }}$

The expression for Fundamental Frequency

$f=\frac{\nu }{2l}$

$f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}$

$f=\frac{1}{2l}\times \sqrt{\frac{T}{\frac{m}{l}}}$

$f=\frac{1}{2l}\times \sqrt{\frac{Tl}{\rho V}}$

Squaring

$f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}$

$T=4\rho \cdot A\cdot l^2\cdot f^2$

$T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2$

$T=7.2\times 10^{-4} N$

8 0

3 years ago

Explain how a generator creates electricity.

vlada-n [284]

Generators don't actually create electricity. Instead, they convert mechanical or chemical energy into electrical energy. They do this by capturing the power of motion and turning it into electrical energy by forcing electrons from the external source through an electrical circuit. hope it helps man

5 0

3 years ago

Read 2 more answers

Hey if your answering this can you just answer the number 14 I already got the rest<br> thanks

Scrat [10]

Layer a is the answer

7 0

2 years ago

Read 2 more answers

A robot is exploring charon, the dwarf planet pluto's largest moon. Gravity on charon is 0.278 meters per second [m divided b

DochEvi [55]

In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up

First we will find the mass of the stone

As it is given that stone is spherical in shape so first we will find its volume

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi *(\frac{0.06}{2})^3$

$V = 1.13 * 10^{-4} m^3$

Now it is given that it's specific gravity is 10.8

So density of rock is

$\rho = 10.8 * 10^3 kg/m^3$

mass of the stone will be

$m = \rho V$

$m = 10.8* 10^3 * 1.13 * 10^{-4}$

$m = 1.22 kg$

now change in potential energy is given as

$\Delta U = mgH$

here

g = gravity on planet = 0.278 m/s^2

H = height lifted upwards = 15 cm

$\Delta U = 1.22* 0.278 * 0.15$

$\Delta U = 0.051 J$

Now energy supplied by internal circuit of robot is given by

$E = Vit$

V = voltage supplied = 10 V

i = current = 1.83 mA

t = time = 12 s

$E = 10* 1.83 * 10^{-3} * 12$

$E = 0.22 J$

Now efficiency is defined as the ratio of output work with given amount of energy used

$\eta = \frac{\Delta U}{E}*100$

$\eta = \frac{0.051}{0.22} = 0.23$

so efficiency will be 23 %

5 0

2 years ago