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jonny [76]
3 years ago
6

Dangerous of overloading in a circuit,​

Engineering
1 answer:
Leto [7]3 years ago
6 0

Answer:

Exceeding the rated load for the circuit wiring causes the circuit breaker to trip, shutting off the power to the entire circuit. If there were no breaker in the circuit, an overload would cause the circuit wiring to overheat, which could melt the wire insulation and lead to a fire

Explanation:

need thanks and make me brainiest if it helps you

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The current through a 10-mH inductor is 10e−t∕2 A. Find the voltage and the power at t = 8 s.
NNADVOKAT [17]

Answer:

voltage = -0.01116V

power = -0.0249W

Explanation:

The voltage v(t) across an inductor is given by;

v(t) = L\frac{di(t)}{dt}             -----------(i)

Where;

L = inductance of the inductor

i(t) = current through the inductor at a given time

t = time for the flow of current

From the question:

i(t) = 10e^{-t/2}A

L = 10mH = 10 x 10⁻³H

Substitute these values into equation (i) as follows;

v(t) = (10*10^{-3})\frac{d(10e^{-t/2})}{dt}

Solve the differential

v(t) = (10*10^{-3})\frac{-1*10}{2} (e^{-t/2})

v(t) = -0.05 e^{-t/2}

At t = 8s

v(t) = v(8) = -0.05 e^{-8/2}

v(t) = v(8) = -0.05 e^{-4}

v(t) = -0.05 x 0.223

v(t) = -0.01116V

(b) To get the power, we use the following relation:

p(t) = i(t) x v(t)

Power at t = 8

p(8) = i(8) x v(8)

i(8) = i(t = 8) = 10e^{-8/2}

i(8) = 10e^{-4}

i(8) = 10 x 0.223

i(8) = 2.23

Therefore,

p(8) = 2.23 x -0.01116

p(8) = -0.0249W

7 0
3 years ago
Read 2 more answers
The difference in quantity between the add and full marks on an engine oil dipstick is typically
svetoff [14.1K]

Answer:

1 quart (0.9 liters).

Explanation:

A proper inspection of various systems and components in a vehicle at regular intervals is very important and necessary because it helps to ensure that the vehicle is in a safe and reliable condition.

Generally, these inspection includes tyres, lighting systems, fan belts, shock absorbers, fluid (oil and water) level, etc. If any fault or concern is detected in the course of an inspection, it should be noted for quick repair or servicing by an expert technician.

All automobile engine requires an adequate amount of engine oil as a lubricant so as to mitigate friction and enhance proper functionality of the vehicle. Thus, the proper functionality of an engine is largely dependent on the level of the engine oil; it shouldn't be too low or high.

Basically, the engine oil should be checked at regular intervals (periodically) and should be on the level indicated or chosen by the manufacturer of the vehicle.

A dipstick is designed to be used for checking the engine oil level in a vehicle and it is marked with lines indicating minimum and maximum, low and high or add and full.

The difference in quantity between the add and full marks on an engine oil dipstick is typically 1 quart (0.9 liters).

5 0
2 years ago
1. A flywheel is suspended by resting the inside of the rim on a horizontal knife edge so that the wheel can swing in a vertical
sammy [17]

Answer: A fly wheel having a mass of 30kg was allowed to swing as pendulum about a knife edge at inner side of the rim as shown in figure.

Explanation:

8 0
3 years ago
Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/
kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k

Obtain the following properties at 10kPa from the table "saturated water"

h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K

Calculate the enthalpy at exit of the turbine using the energy balance equation.

\frac{dE}{dt}=Q-W+m(h_1-h_2)

Since, the process is isentropic process Q=0

0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg

Use the isentropic relations:

s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87

Calculate the enthalpy at isentropic state 2s.

h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg

a.)

Calculate the isentropic turbine efficiency.

\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%

b.)

Find the quality of the water at state 2

since h_f at 10KPa <h_2<h_g at 10KPa

Therefore, state 2 is in two-phase region.

h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9

Calculate the entropy at state 2.

s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K

Calculate the rate of entropy production.

S=\frac{Q}{T}+m(s_2-s_1)

since, Q = 0

S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k

6 0
3 years ago
Can i use two shunts and one meter
Lyrx [107]

Answer:

no

Explanation:

6 0
3 years ago
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