An electric bulb rated 100 W, 100 V has to be

operated aross 141.4 V, 50 Hz A.C. supply. The

capacitance of the capacitor which has to be

connected in series with bulb so that bulb will

glow with full intensity is [NCERT Pg. 251]

Answers

Answer 1

Answer:

The capacitance of the capacitor is 31.84 μF.

Explanation:

Given;

power rating of the bulb, P = 100 W

voltage rating of the bulb, Vr = 100 V

operating voltage of the bulb, V= 141.4 V

frequency of the AC = 50 Hz

P = IV = 100 W

V = 100 V

I =

Ic = 1 A

The voltage across the capacitor is given by;

[tex]V_c = \sqrt{V^2 - V_R^2} \\\\V_c = \sqrt{141.4^2 - 100^2} \\\\V_c =99.97 \ V[/tex]

[tex]V_c = I_cX_c\\\\V_c = I_C* \frac{1}{2\pi fC}\\\\ 99.97 = 1 * \frac{1}{2\pi *50 *C}\\\\ C=\frac{1}{2\pi *50*99.97}\\\\ C = 31.84*10^{-6} \ F\\\\C = 31.84 \ \mu F[/tex]

Therefore, the capacitance of the capacitor is 31.84 μF.


Related Questions

While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?

Answers

Analysing the question:

Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s

We are given:

height of the tower (h) = 66 m

mass of the stone (m) = 0.5 kg

initial velocity of the stone (u) = 0 m/s

time taken by the stone to reach the ground (t) = t seconds

acceleration due to gravity = 10 m/s²

** Neglecting air resistance**

Finding the time taken by the stone to reach the ground:

from the second equation of motion

h = ut + 1/2at²

replacing the variables

66 = (0)(t) + 1/2 (10)(t)²

66 = 5t²

t² = 13.2

t = 3.6 seconds

I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds

but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved

A battery is used to charge a parallel-plate capacitor, after which it is disconnected. Then the plates are pulled apart to twice their original separation. This process will double the: __________A. capacitance
B. surface charge density on each plate
C. stored energy
D. electricfield between the two places
E. charge on each plate"

Answers

Answer: C.

Explanation:

For a parallel-plate capacitor where the distance between the plates is d.

The capacitance is:

C = e*A/d

You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.

Now, the stored energy can be written as:

E = (1/2)*Q^2/C

Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:

E = (1/2)*Q^2*d/(e*A)

e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.

Then we can define:

K = (1/2)*Q^2/(e*A)

And now we can write the energy as:

E = K*d

Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.

If the plates are pulled apart to twice their original separation, then this will double the stored energy. Hence, option (C) is correct.

The given problem is based on the concept of parallel plat capacitor. For a parallel-plate capacitor where the distance between the plates is d.

The capacitance is:

C = e*A/d

here.

e is the permittivity of free space.

Since, the distance is inversely proportional then if we double the distance, the capacitance halves.  Now, the stored energy can be given as,

E = (1/2)*Q^2/C

here,

Q is the charge stored in the capacitor.

Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:

E = (1/2)*Q^2*d/(e*A)

e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.

Then we can define:

K = (1/2)*Q^2/(e*A)

And now we can write the energy as:

E = K*d

So, the energy is proportional to the distance between the plates.

Thus, we can conclude that if the plates are pulled apart to twice their original separation, then this will double the stored energy. Hence, option (C) is correct.

Learn more about the energy stored in a capacitor here:

https://brainly.com/question/3611251

A small compass is held horizontally, the center of its needle has a distance of 0.270 m directly north
of a long wire that is perpendicular to the Earth's surface. When there is no current in the wire, the
compass needle points due north, which is the direction of the horizontal component of the Earth's
magnetic field at that location. This component is parallel to the Earth's surface. When the current in
the wire is 26.3 A, the needle points 22.9∘ east of north.
(a) Does the current in the wire flow toward or away from the Earth's surface? ( 2 marks)
(b) What is the magnitude of the horizontal component of the Earth's magnetic field at the location of
the compass? (3 marks)

Answers

Answer:

Explanation:

The needle is showing north south direction . when current starts flowing in the wire which is held vertical to the ground , it deflects towards east .

a )

Therefore a magnetic field towards east has been created . It is possible only if current flows towards the surface in the vertical wire .

b )

magnetic field created at the magnetic needle B = 10⁻⁷ x  2I / d where I is current and d is distance .

B = 10⁻⁷ x  2 x 26.3  / .27

= 194.81 x 10⁻⁷ T

angle of deflection of solenoid = 22.9°

Tan 22.9 = B /H

.422 = 194.81 x 10⁻⁷ / H

H = 461.63 x 10⁻⁷ T

= .46 x 10⁻⁴ T .

A) The current in the wire flows towards the Earth's surface

B) The magnitude of the horizontal component of the Earth's magnetic field is :   0.46 x 10⁻⁴ T

A) The compass needle held horizontally points in a North-south direction of the earth and also deflects eastwards when current is allowed to flow through it. The deflection of the needle indicates the presence/generation of a magnetic field on the earth surface. which is facilitated by the flow of the current in the wire towards the Earth's surface

B) Determine The magnitude of the horizontal component of the Earth's magnetic field

B ( magnetic field ) = 10⁻⁷ * 2I / d ---- ( 1 )

where : l = 26.3 A,   d = 0.27 m

Back to equation ( 1 )

B = 10⁻⁷ * 2 * 26.3 / 0.27

  = 194.81 * 10⁻⁷ T

Final step : Calculate the magnitude of horizontal component  ( H )

Tan ∅ = B / H ---- ( 2 )

where : ∅ ( angle of deflection ) = 22.9°

∴ H = B / Tan ( 22.9° )

      = (  194.81 * 10⁻⁷ ) / 0.422

      = 0.46 x 10⁻⁴ T

Hence we can conclude that The current in the wire flows towards the Earth's surface and  The magnitude of the horizontal component of the Earth's magnetic field is :   0.46 x 10⁻⁴ T

Learn more about Earth magnetic field : https://brainly.com/question/115445

A negative charge -Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. Is any excess charge induced on the inner surface of the metal? Is there any excess charge on the outside surface of the metal? Why or why not? Would someone outside the solid measure an electric field due to the charge -Q? Is it reasonable to say that the grounded conductor has shielded the region outside the conductor from the effects of the charge -Q? In principle, could the same thing be done for gravity? Why or why not?

Answers

Answer:

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) E=0

Explanation:

Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.

With this analysis we can answer the specific questions

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside

d) If it is very reasonable and this system configuration is called a Faraday Cage

e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.

If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 20.0 km? Take the mass of the sun to be 1.99×10^30, the gravitational constant to be G = 6.67×10^−11Nm^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.p

Answers

Answer:

8.93*10^13 N.

Explanation:

Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:

       [tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]

where, m1 = mass of  the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.Replacing by the values, we get:

       [tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]

Fg = 8.93*10^13 N.

help me get the answer in Physical Science.

Answers

Answer:

lithium

Explanation:

I took physical science 2 years ago and passed with an A

A motorboat is a lot heavier than a pebble. Why does the boat float?

Answers

Answer:

The boat has more buoyancy

Explanation:

A particle moves along a path described by y=Ax^3 ​​ and x = Bt, where tt is time. What are the units of A and B?

Answers

Answer:

In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.

Explanation:

From Dimensional Analysis we understand that [tex]x[/tex] and [tex]y[/tex] have length units ([tex][l][/tex]) and [tex]t[/tex] have time units ([tex][t][/tex]). Then, we get that:

[tex][l] = A\cdot [l]^{3}[/tex] (Eq. 1)

[tex][l] = B\cdot [t][/tex] (Eq. 2)

Now we finally clear each constant:

[tex]A = \frac{[l]}{[l]^{3}}[/tex]

[tex]A = \frac{1}{[l]^{2}}[/tex]

[tex]B = \frac{[l]}{[t]}[/tex]

In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.

In the absence of a gravitational field, you could determine the mass of an object (of unknown composition) by:
A) applying a known force and measuring it's acceleration.
B) measuring the volume.
C) weighing it.

Answers

Answer:

A) By applying a known force, and measuring it's acceleration.

Explanation:

This is actually something that astronauts do in space as a mathmatical exercise when calculating the mass of an object since F = m × a.

Once the force, and acceleration are applied, the only unknown is the mass which can be solved by dividing force over acceleration. This is because inertial mass is equal to gravitational mass.

what is the force produced on a body of 30kg mass when a body moving with the velocity of 26km/hr is acceleted to gain the velocity of 54 km/hr in 4 sec​

Answers

Answer:

F = 58.35 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration. But first we must use the following equation of kinematics.

We have to convert speeds from kilometers per hour to meters per second

[tex]\frac{26km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=\frac{7.22m}{s} \\\frac{54km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex]

[tex]v_{f}=v_{o}+(a*t) \\[/tex]

where:

Vf = final velocity = 15 [m/s]

Vi = initial velocity = 7.22 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Note: the positive sign of the above equation is because the car increases its speed

15 = 7.22 + (a*4)

a = 1.945 [m/s^2]

Now we can use the Newton's second law:

F = m*a

F = 30*1.945

F = 58.35 [N]

It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Answers

Answer:

102 m

Explanation:

Given that It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Let the stopping distance be equal to S.

According to the definition of speed,

Speed = distance / time.

make time the subject of the formula

Time = distance / speed

then, the equivalent time is:

48.96 / 12 = S / 25

Cross multiply

12S = 48.96 x 25

12S = 1224

S = 1224 / 12

S = 102 m

Therefore, the stopping distance is 102 m

You release a ball from rest at the top of a ramp. 6 s later it is moving at 4.0
m/s. What is the acceleration? (in meters per second squared) *
Your answer

Answers

[tex]a = \frac{vf - vi}{t} [/tex]

here initial velocity vi=0 as ball release from rest

the final velocity is vf=4.0

time is t=6

so putting all these values in above equation

[tex]a = \frac{ 4.0- 0}{6} [/tex]

[tex]a = 0.6667m \s {}^{2} [/tex]

How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential difference, it is connected to the power supply and resistor and onehalf the characteristic time has passed (i.e. t= T(tau)/2)?

Answers

Answer:

The voltage is   [tex]V =   0.993V_b[/tex]

Explanation:

From the question we are told that

   The time that has passed is  [tex]t = \frac{\tau}{2}[/tex]

 Here [tex]\tau[/tex] is know as the time constant

    The voltage of the  power source is   [tex]V_b[/tex]

Generally the voltage equation for charging a capacitor is mathematically represented as

       [tex]V =  V_b  [1 - e^{- \frac{t}{\tau} }][/tex]

=>   [tex]V =  V_b  [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }][/tex]

=>   [tex]V =  V_b  [1 - e^{- \frac{\tau}{2\tau} }][/tex]

=>   [tex]V =  V_b  [1 - e^{- \frac{1}{2} }][/tex]

=>   [tex]V =   0.993V_b[/tex]    

An FM radio station, 20 miles away, broadcast at a 93.4 MHz frequency(a) What is the wavelength of the radio wave associated with this signal ?(b) How long does it take for the signal to reach your radio from the station ?

Answers

Answer:

(a) Wavelength = 3.21 m (b) Time = [tex]1.07\times 10^{-4}\ s[/tex]

Explanation:

Given that,

The frequency of FM radio station, f = 93.4 MHz

(a) We need to find the wavelength of the radio wave associated with this signal. The relation between wavelength and frequency is given by :

[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{93.4\times 10^6}\\\\\lambda=3.21\ m[/tex]

(b) It is given that, an FM radio station, 20 miles away. Let t is time taken for signal to reach your radio from the station. So,

[tex]t=\dfrac{d}{c}\\\\t=\dfrac{20\times 1609.34}{3\times 10^8}\\\\t=1.07\times 10^{-4}\ s[/tex]

Hence, this is the required solution.

21. Prediction: If you were to measure the current at points A, B and C, how do you think the values would compare? Why? 22. Prediction: If you were to measure the potential differences across these bulbs (what the voltmeter measures) how do you think the values will compare to each other and to the potential difference across the battery pack or the power supply? Why?

Answers

Answer:

hello your question is incomplete attached below is the complete question

21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C

22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)

hence the voltage in the battery will be equal to the voltage across each bulb

Explanation:

The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C

The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)

hence the voltage in the battery will be equal to the voltage across each bulb

If it takes you 5 minutes to dry your hair using a 1200-W hairdryer plugged into a 120-V power outlet, how many Coulombs of charge pass through your hair dryer

Answers

Answer:

The charge pass through your hair dryer is 3000 C.

Explanation:

Given that,

Power = 1200 W

Voltage = 120 V

Flow time = 5 min

We need to calculate the current

Using formula of power

[tex]P=VI[/tex]

[tex]I=\dfrac{P}{V}[/tex]

Put the value into the formula

[tex]I=\dfrac{1200}{120}[/tex]

[tex]I=10\ A[/tex]

We need to calculate the charge pass through your hair dryer

Using formula of current

[tex]I=\dfrac{Q}{t}[/tex]

[tex]Q=It[/tex]

Put the value into the formula

[tex]Q=10\times5\times60[/tex]

[tex]Q=3000\ C[/tex]

Hence, The charge pass through your hair dryer is 3000 C.

the peripheral nervous system is responsible for both sending and receiving signals to and from the brain

Answers

Answer:

its true trust me

Explanation:

Answer: true

Explanation: edge

the diagram shows a contour map. letter a through k are reference points on the map. which points are located at the same elevation above sea level?

Answers

Answer:

K and I

Explanation:

Contour maps use lines that represent spaces in a map that have the same elevation, this means that all the lines should be continuous and closed, in this case, we are not able to see the full extent of most of the lines, but since the points are located in different lines we can assume that they are at different heights, so since only point K and point I are on the same line, we know that these two points are at the same height.


Weight of a body becomes greater at the pole than that at the equator . why ?

Answers

Answer:

Because the Earth rotates it is wider around at the equator than around the pole.
The distance from the Pole to the centre is smaller than the distance from the equator to the centre of the Earth. The weight decreases the further away from the centre of the Earth.

Sometimes we will want to write vectors in terms of a coordinate grid. To show a vector points
horizontally (along the x-axis), place an x after the magnitude of the vector. To show a vector point
vertically (along the y-axis), place a y after the magnitude.
4) Using the notation above,
i. How would you write d1?
ii. How would you write d2?
iii. How would you write dtotal?
d1=(0,5)
d2=(5,5)

Answers

Answer:

III) [tex]d_{1}+ d_{2}=d_{t}[/tex]

Explanation:

I) coordinate (0,5) is the head for [tex]d_{1}[/tex] I will put the tail coordinate as (0,0) but it could be any other number in the x just not in the 5  with the the y being any other value.

II) coordinate (5,5) is the head for [tex]d_{2}[/tex] the tail needs to be in the head of [tex]d_{1}[/tex] being (0,5)

III) coordinates for [tex]d_{t}[/tex] is connecting the tail from [tex]d_{1}[/tex] and the head of [tex]d_{2}[/tex] making it (0,0)[tex](tail)[/tex] and (0,5)[tex](head)[/tex] and is written as [tex]d_{1}+ d_{2}=d_{t}[/tex]

(i) using coordinate grid notation to represent d₁, d₁ = 5y

(ii) using coordinate grid notation  to represent d₂, d₂ = 5x + 5y

(ii) The sum of d₁ and d₂ is written as 5x + 10y

In order to show the horizontal direction of a vector, we will place x after the magnitude of the vector.

Also, to show the vertical direction of a vector, we will place a y after the magnitude of the vector.

(i) Using coordinate grid to represent d = (0, 5)

[tex]d_1 = 0(x) + 5(y)\\\\d_1 = 5y[/tex]

(ii) Using coordinate grid to represent d₂ = (5, 5)

[tex]d_2 = 5x + 5y[/tex]

(iii) The total vector is written as;

[tex]d_1 + d_2 = 5y + (5x + 5y)\\\\d_1 + d_2 = 5y + 5x + 5y\\\\d_1 + d_2 = 5x + 10y[/tex]

Learn more here: https://brainly.com/question/17212749

ionic bonds form when electrons?

Answers

Answer:

when the electron transferred permanently to another atom

7. A 1,500-N force is applied to a 1,000-kg car. What is the car's acceleration?

Answers

Answer:

1.5m/s^2

Explanation:

Answer:

1.5 m/s2. accerelation =force ÷mass

What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75

Answers

Answer:

1800 m/s

Explanation:

The equation is v = fλ

λ= 0.75

f = 2400 Hz

V = 2400 × 0.75

V = 1800 m/s

[ you did not give units for wavelength, I assumed it would be m/s]

A plane flying horizontally at a speed of 40.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?

Answers

Answer:

Package 1 will land at 228.0 m, package 2 will land at 308.0 m, and the distance between them is 80.0 m.

 

Explanation:

To find the distance at which the first package will land we need to calculate the time:

[tex] Y_{f} = Y_{0} + V_{0y}t - \frac{1}{2}gt^{2} [/tex]

Where:

Y(f) is the final position = 0

Y(0) is the initial position = 160 m

V(0y) is initial speed in "y" direction = 0

g is the gravity = 9.81 m/s²

t is the time=?                                          

[tex] 0 = 160 m + 0t - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]

[tex] t = \sqrt{\frac{2*160 m}{9.81 m/s^{2}}} = 5.7 s [/tex]

Now we can find the distance of the first package:

[tex] X_{1} = V_{0x}*t = 40.0 m/s*5.7 s = 228.0 m [/tex]

Then, after 2 seconds the distance traveled by plane is (from the initial position):

[tex] X_{p} = V_{0x}*t = 40.0 m/s*2 s = 80.0 m [/tex]

Now, the distance of the second package is:

[tex] X _{2} = X_{1} + X_{p} = 228.0 m + 80.0 m = 308.0 m [/tex]

The distance between the packages is:

[tex] X = X_{2} - X_{1} = 308.0 - 228.0 m = 80.0 m [/tex]

Therefore, package 1 will land at 228.0 m, package 2 will land at 308.0 m and the distance between them is 80.0 m.

I hope it helps you!

During a thunderstorm the electric field at a certain point in the earth's atmosphere is 1.07 105 N/C, directed upward. Find the acceleration of a small piece of ice of mass 1.08 10-4 g, carrying a charge of 1.05 10-11 C.

Answers

Answer:

The acceleration of a small piece of ice is 10.40 m/s².

Explanation:

The electric force is given by:

[tex]F = Eq[/tex]

Where:    

E is the electric field = 1.07x10⁵ N/C

q is the charge = 1.05x10⁻¹¹ C          

The electric force is equal to Newton's second law:

[tex] Eq = ma [/tex]

Where:            

m is the mass = 1.08x10⁻⁴ g = 1.08x10⁻⁷ kg

a is the acceleration

Hence, the acceleration is:

[tex] a = \frac{Eq}{m} = \frac{1.07 \cdot 10^{5} N/C*1.05 \cdot 10^{-11} C}{1.08 \cdot 10^{-7} kg} = 10.40 m/s^{2} [/tex]

Therefore, the acceleration of a small piece of ice is 10.40 m/s².

I hope it helps you!                    

Jumping on a trampoline cause you to fly up in the air. What type of newton’s law is it ?

Answers

Answer:

The Third law

Explanation:

For every action there is an equal and opposite reaction.

Answer:

First Law

Explanation:

An object at rest (not moving) will stay at rest unless an unbalanced force acts on it.

An object in motion will stay in motion (in a straight line and at a constant speed) unless an unbalanced force acts on it.

You jump down on a trampoline and fly up in the air as a result.

Other Questions
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