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Y_Kistochka [10]
3 years ago
5

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

Engineering
1 answer:
GrogVix [38]3 years ago
5 0

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1

a) 0.01 μF

b) 0.58 μF

c) 0.060 μF

d) 0.8 μF

Answer:

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

Explanation:

Please refer to the attached Figure 12-1 where three capacitors are connected in series.

We are asked to find out the equivalent capacitance of this circuit.

Recall that the equivalent capacitance in series is given by

$ \frac{1}{C_{eq}} =  \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} $

Where C₁, C₂, and C₃ are the individual capacitance connected in series.

C₁ = 0.1 μF

C₂ = 0.22 μF

C₃ = 0.47 μF

So the equivalent capacitance is

$ \frac{1}{C_{eq}} =  \frac{1}{0.1} + \frac{1}{0.22} + \frac{1}{0.47} $

$ \frac{1}{C_{eq}} =  \frac{8620}{517}  $

$ C_{eq} =  \frac{517}{8620}  $

$ C_{eq} =  0.0599  $

Rounding off yields

$ C_{eq} =  0.060 \: \mu F $

The equivalent capacitance of the three series capacitors in Figure 12-1 is 0.060 μF

Therefore, the correct option is (c)

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Sladkaya [172]

Answer:

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Explanation:

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5 0
3 years ago
A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po
Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: 4.3\cdot 10^{-15} J

Explanation:

a)

The electric potential at a distance r from a single-point charge is given by:

V(r)=\frac{kq}{r}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

q_1=+2.00\mu C=+2.00\cdot 10^{-6}C

and is located at the origin (x=0, y=0)

Charge 2 is

q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

r_{1A}=0.40 m

So the potential due to charge 2 is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V

The distance of point A from charge 2 is

r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m

So the potential due to charge 1 is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V

Therefore, the net potential at point A is

V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V

b)

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m

So the potential due to charge 1 at point B is

V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V

The distance of charge 2 from point B is

r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m

So the potential due to charge 2 at point B is

V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V

Therefore, the net potential at point B is

V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V

So the potential difference is

V_B-V_A=-29,500 V-(-2700 V)=-26,800 V

c)

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

W=q\Delta V

where

q is the charge of the particle

\Delta V is the potential difference

In this problem, we have:

q=-1.6\cdot 10^{-19}C is the charge of the electron

\Delta V=-26,800 V is the potential difference

Therefore, the work required on the electron is

W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J

4 0
3 years ago
What are the controlling LRFD load combinations for dead and floor live load?
yuradex [85]

Answer:

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

Explanation:

Load and Resistance Factor Design

there are 7 basic load combination of LRFD that is

1) 1.4(D + F)

2) 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

3) 1.2D + 1.6(Lr or S or R) + ((0.5 or 1.0)*L or 0.8W)

4) 1.2D + 1.6W + (0.5 or 1.0)*L + 0.5(Lr or S or R)

5) 1.2D + 1.0E + (0.5 or 1.0)*L + 0.2S

6) 0.9D + 1.6W + 1.6H

7) 0.9D + 1.0E + 1.6H

and

here load factor for L given ( * ) mean it is  permitted = 0.5 for occupancies when live load is less than or equal to 100 psf

here

D is dead load and L is live load

E is earth quake load and S is snow load

W is wind load and R is rain load

Lr is roof live load

3 0
3 years ago
Air flows through a 0.25-m-diameter duct. At the inlet the velocity is 300 m/s, and the stagnation temperature is 90°C. If the M
Naddika [18.5K]

Answer:

a. 318.2k

b. 45.2kj

Explanation:

Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.

See attachment for detailed analysis

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3 years ago
The Torricelli's theorem states that the (velocity—pressure-density) of liquid flowing out of an orifice is proportional to the
Sergeeva-Olga [200]

Answer:

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Explanation:

Torricelli's theorem states that

v_{exit}=\sqrt{2gh}

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v_{exit} is the velocity with which the fluid leaves orifice

h is the head under which the flow occurs.

Thus we can compare the given options to arrive at the correct answer

Velocity is proportional to square root of head under which the flow occurs.

4 0
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