Cast iron has about how much carbon content?

A coil having resistance of 7 ohms and inductance of 31.8 mh is connected to 230v,50hz supply.calculate 1. The circuit current 2

lora16 [44]

(1) The current in the circuit is 18.87 A,

(2) The phase angle is 54.97°

(3) The power factor is 0.574

(4) The power consumed is 2491.2 W

(1) To calculate the current in the circuit, first, we need to find the overall impedance of the circuit.

We can calculate the overall impendence of the circuit using the formula below.

Z = √[R²+(2πfL)²]........................ Equation 1

Where:

R = resistance of the coil
f = Frequency
L = Inductance of the coil
Z = Overall impedance of the circuit

From the question,

Given:

R = 7 ohms
L = 31.8 mH = 0.0318 H
f = 50 Hz
π = 3.14

Substitute these values into equation 1

Z = √[7²+(2×3.14×50×0.0318)²]
Z = √(49+99.7)
Z = √(148.7)
Z = 12.19 ohms.

Therefore we use the formula below to calculate the current in the circuit.

I = V/Z.................. Equation 2

Where:

V = Voltage
I = current in the circuit.

Given:

V = 230 V.

Substitute into equation 2

I = 230/12.19
I = 18.87 A

(2) To calculate the phase angle, we use the formula below.

∅ = tan⁻¹(2πfL/R)............... Equation 3

Where:

∅ = Phase angle.

Substitute into equation 3

∅ = tan⁻¹(2×3.14×50×0.0318/7)
∅ = tan⁻¹(9.9852/7)
∅ = tan⁻¹(1.426)
∅ = 54.97°

(3) To calculate the power factor, we use the formula below.

pf = cos∅............ Equation 4

Where:

pf = power factor.

Substitute the value of ∅ into equation 4

pf = cos(54.97°)
pf = 0.574.

(4) And Finally to calculate the power consumed we use the formula below.

P = V×I×pf................ Equation 5

Where:

P = The power consumed

Substitute the values into equation 5

P = 230(18.87)(0.574)
P = 2491.22 W

Hence, (1) The current in the circuit is 18.87 A, (2) The phase angle is 54.97° (3) The power factor is 0.574 (4) The power consumed is 2491.2 W

Learn more about Impedance here: brainly.com/question/13134405

5 0

2 years ago

The depletion in the Ozone layer is caused by:

andrew11 [14]

D. Chlorofluorocarbon

This is a man-made carbon that causes the gradual thinning(deception) in the earth’s Ozone layer.

Hope this helped and pls mark as brainliest!

~ Luna

3 0

2 years ago

Read 2 more answers

What is made in heaven?

kramer

Answer:

Babies come from heaven didn't you know?

3 0

2 years ago

What are three common hazardous

Delicious77 [7]

Answer:

SNOW SQUALLS - Squalls are substantial snow showers that cause white-out conditions. A couple of squalls are conceivable at the beginning of today through mid-evening. The most elevated possibility is north of the expressway.
COLD - Because of away from and a new cover of day off, have plunged to the youngsters and 20s. There are frigid untreated surfaces.
WINDS – One can expect supported breezes 15-25 mph this evening, with blasts to 40 mph. Solid breezes, alongside temperatures during the 30s, will create wind chills during the 20s.
FOG - Perceivability is improving, however parts of Lancaster, Lebanon, Schuylkill, Berks, and Dauphin Provinces have decreased perceivability. With temperatures beneath freezing, mist can stick to the streets and produce frigid conditions.

Explanation:

An effect day in Pennsylvania is a day that highlights climate that will probably upset your typical every day calendar or schedule. At the point when the News 8 Tempest Group accept climate arrives at this limit, a yellow triangle with a shout imprint will show up on the figure demonstrating the climate will cause an effect. You may likewise observe the symbol utilized during a day-part figure when the News 8 Tempest Group accepts certain times may have a greater amount of an effect than others. An effect day's climate isn't viewed as outrageous or serious, perilous climate.

Light downpour, day off, ice during a morning or night drive would be instances of effect climate. Most Pennsylvania workers can deal with driving in light downpour, day off, ice, however light downpour, day off, ice at the pinnacle of busy time can upset travel or cause mishaps because of the high volume of drivers. Thick mist over the Valley which could likewise modify a drive would be delegated an effect day.

Moderate cold throughout the fall, winter, spring months, or warmth in the mid year months that may constrain open air presentation to the components would get this assignment.

The Susquehanna Valley is a significant agrarian community for the locale. Pre-summer or late-summer ice or freezes that could harm or murder vegetation would fit into this class.

Days that component consume bans from delayed droughts combined with solid breeze blasts likewise would be considered to conceivably make an effect a day ahead.

4 0

3 years ago

1. 6.1 PSPICEMULTISIM The current in a 50μH inductor is known to be iL=18te−10tAfor t≥0. 1. Find the voltage across the inductor

Andre45 [30]

Answer:

a. Voltage across the inductor for t > 0 is 0.9e^-10t(1-10t)

b. Power = -59.3μW

c. Inductor is delivering power.

d. Energy = 5934.3nJ

e. Time = 100ms; Energy = 1095941.025nJ

Explanation:

Given

Current; iL=18te^(−10t)A or t≥0.

L.= inductor = 50μH

a. The voltage, V across the inductor for t>0 is calculated as follows;

V = L(di/dt)

Where L = 50μH

di/dt = 18(e^-10t + (-10)te^-10t)

di/dt = 18e^-10t(1 - 10t)

Substitute 50μH for L and 18e^-10t(1 - 10t) for di/dt in V = L(di/dt)

V = 50μH * 18e^-10t(1 - 10t)

V = 50 * 10^-6(18e^-10t(1 - 10t))

V = 0.9e^-10t(1-10t)

Hence, the voltage across the inductor for t > 0 is 0.9e^-10t(1-10t)

b. Find the power (in microwatts) at the terminals of the inductor when t=200 ms.

Given that t = 200ms = 200 * 10^-3s = 0.2s

Power, p is calculated using the following formula;

p = Li(di/dt)

p = 50 * 10^-6(18te^-10t)18e^-10t(1-10t)

p = 50 * 10^-6 * (18 * 0.2 * e^-(10*0.2)) * (18 * e^(-10 * 0.2) * (1-10*0.2)

p = -5.93E5W

p = -59.3μW

c. Is the inductor absorbing or delivering power at 200 ms?

Because of the negative sign, the inductor is delivering power.

d. Find the energy (in microjoules) stored in the inductor at 200 ms.

Energy is calculated as ½Li²

= ½ * 50 * 10^-6 * (18te^-10t)²

= ½ * 50 * 10^-6 * (18 * 0.2 * e ^ (-10 * 0.2))²

= 0.0000059342669999498J

= 5934.3nJ

e. Find the maximum energy (in microjoules) stored in the inductor and the time (in milliseconds) when it occurs.

Calculating the derivation in (a)

di/dt = 0

18e^-10t(1-10t) = 0

1 - 10t = 0

-10t = -1

t = 1/10

t = 100ms

To calculate the energy, first we need to calculate the current

I(t=100) = 18 * 0.1 * e^(-10(0.1)

I = 0.662182994108596

I = 6621.82mA

The energy is calculated as follows;

w = ½ * 50 * 10^-6 * (6.621)²

w = 0.001095941025

w = 1095941.025nJ

8 0

3 years ago