Ask question

Ask question

All categories

2 years ago

11

For the circuit, suppose C=10µF, R1=1000Ω, R2=3000Ω, R3=4000Ω and ls=1mA. The switch closes at t=0s.1) What is the value of Vc (

in volts) just prior to the switch closing? Assume that the switch had been open for a long time. 2) For the circuit above, what is the value of Vc after the switch has been closed for a long time?
3) What is the time constant of the circuit (in seconds)? Enter the answer below without units.
4) What is the value of Vc at t = 2msec (in volts).

1 answer:

umka21 [38]2 years ago

5 0

Answer:

1.) Vc = 1V

2.) Vc = 2.7V

3.) Time constant = 0.03

4.) V = 2.53V

Explanation:

1.) The value of Vc (in volts) just prior to the switch closing

The starting current = 1mA

With resistance R1 = 1000 ohms

By using ohms law

V = IR

Vc = 1 × 10^-3 × 1000

Vc = 1 volt.

2.) The value of Vc after the switch has been closed for a long time.

R2 and R3 are in parallel to each other. Both will be in series with R1

The equivalent resistance R will be

R = (R2 × R3)/R2R3 + R1

Where

R1 = 1000Ω,

R2 = 3000Ω,

R3 = 4000Ω

R = (4000×3000)/(4000+3000) + 1000

R = 12000000/7000 + 1000

R = 1714.3 + 1000

R = 2714.3 ohms

By using ohms law again

V = IR

Vc = 1 × 10^-3 × 2714.3

Vc = 2.7 volts

3.) The time constant = CR

Time constant = 10 × 10^-6 × 2714.3

Time constant = 0.027

Time constant = 0.03 approximately

4.) The value of Vc at t = 2msec (in volts). Can be calculated by using the formula

V = Vce^-t/CR

Where

Vc = 2.7v

t = 2msec

CR = 0.03

Substitute all the parameters into the formula

V = 2.7 × e^-( 2×10^-3/0.03)

V = 2.7 × e^-(0.0667)

V = 2.7 × 0.935

V = 2.53 volts

You might be interested in

A newly discovered planet has a mean radius of 7380 km. A vehicle on the planet\'s surface is moving in the same direction as th

Butoxors [25]

Answer:

292796435 seconds ≈ 300 million seconds

Explanation:

First of all, the speed of the car is 121km/h = 33.6111 m/s

The radius of the planet is given to be 7380 km = 7380000 m

From the relationship between linear velocity and angular velocity i.e., v=rw, the angular velocity of the car will be w=v/r = 33.6111/7380000 = 0.000000455 rad/s = 4.55 x 10⁻⁶ rad/sec

If the angular velocity of the vehicle about the planet's center is 9.78 times as large as the angular velocity of the planet then we have

w(vehicle) = 9.78 x w(planet)

w(planet) = w(vehicle)/9.78 = 4.55 x 10⁻⁶ / 9.78 = 4.66 x 10⁻⁷ rad/sec

To find the period of the planet's rotation; we use the equation

w(planet) = 2π÷T

Where w(planet) is the angular velocity of the planet and T is the period

From the equation T = 2π÷w = 2×(22/7) ÷ 4.66 x 10⁻⁷ = 292796435 seconds

Therefore the period of the planet's motion is 292796435 seconds which is approximately 300, 000, 000 (300 million) seconds

8 0

2 years ago

An engineer is designing a process for a new transistor. She uses a vacuum chamber to bombard a thin layer of silicon with ions

adelina 88 [10]

Answer:

$E=5.7\times 10^{-3}\ V/m$

Explanation:

Given that

$m_p=5.18\times 10^{-26}\ kg$

re= 46 cm

Vp= 180 m/s

We know that

$E=\dfrac{\Delta V}{r}$

$\Delta V=\dfrac{1}{4}\dfrac{m_pv_p^2}{e}$

So

$E=\dfrac{1}{4}\dfrac{m_pv_p^2}{e.r_e}$

Now by putting the all given values in the questions

$E=\dfrac{1}{4}\times \dfrac{5.18\times 10^{-26}\times 180^2}{1.6\times 10^{-19}\times 0.46}$

$E=5.7\times 10^{-3}\ V/m$

So the average electric field is $E=5.7\times 10^{-3}\ V/m$ .

6 0

2 years ago

One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its hi

Aneli [31]

Answer:

Explanation:

Given

initial speed(u)=5 m/s

Final speed(v)=4 m/s

Distance traveled=3 m

using equation of motion

$v^2-u^2=2as$

$4^2-5^2=2(a)(3)$

$a=\frac{-3}{2}=-1.5 m/s^2$

after this its final velocity will be zero

$v^2-u^2=2as$

$0^2-4^2=2\times (-1.5)\times s$

s=5.33 m

Total distance=3+5.33=8.33 m

Thus he will not be able to win the game

4 0

2 years ago

A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f

Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

$\mu mg = m\frac{v^2}{r}$

where

$\mu$ is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

$v=\sqrt{\mu gr}$

And by substituting the data of the problem, we find the speed at which the car begins to skid:

$v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s$

7 0

2 years ago

Read 2 more answers

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

adell [148]

Answer:

$F_{x}=2.31N$ to the right.

$F_{y}=2.1N$ to in the upwards direction.

Explanation:

In order to solve this problem, we must first start by drawing a diagram of the situation. (See attached diagram).

So, remember that a force is determined by multiplying the mass of the parcticle by its acceleration:

F=ma

so in order to find the components of the force, we need to start by finding its acceleration.

Acceleration is found by using the following formula:

$a=\frac{V_{f}-V{0}}{t}$

so we can subtract the two vectors, like this:

$a=\frac{(6.00i+4.0j)m/s-1.60i m/s}{8s}$

which yields:

$a=\frac{(4.4i+4.0j)m/s}{8s}$

or:

$a=(0.55i + 0.5j) m/s^{2}$

so now I can find the components of the force:

$F=(4.2kg)(0.55i + 0.5j) m/s^{2}$

which yields:

F=(2.31i+2.1j)N

so the components of the force are:

$F_{x}=2.31N$ to the right.

$F_{y}=2.1N$ to in the upwards direction.

6 0

2 years ago