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A circuit contains four capacitors in parallel (10 F, 3 F, 7 F, and 1 F). What is the equivalent capacitance of this circuit?

Oxana [17]

The equivalent capacitance ( $C_{eq}$ ) of an electrical circuit containing four capacitors which are connected in parallel is equal to: A. 21 F.

<h3>The types of circuit.</h3>

Basically, the components of an electrical circuit can be connected or arranged in two forms and these are;

Series circuit

Parallel circuit

<h3>What is a parallel circuit?</h3>

A parallel circuit can be defined as an electrical circuit with the same potential difference (voltage) across its terminals. This ultimately implies that, the equivalent capacitance ( $C_{eq}$ ) of two (2) capacitors which are connected in parallel is equal to the sum of the individual (each) capacitances.

Mathematically, the equivalent capacitance ( $C_{eq}$ ) of an electrical circuit containing four capacitors which are connected in parallel is given by this formula:

Ceq = C₁ + C₂ + C₃ + C₄

Substituting the given parameters into the formula, we have;

Ceq = 10 F + 3 F + 7 F + 1 F

Equivalent capacitance, Ceq = 21 F.

Read more equivalent capacitance here: brainly.com/question/27548736

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3 0

2 years ago

A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?

jok3333 [9.3K]

Answer:

$S_a_v_e_r_a_g_e=48km/h$

Explanation:

Ok, the average speed can be calculate with the next equation:

$S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}$ (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

$Total\hspace{3}distance=2*d$

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

$Total\hspace{3}time=t1+t2$

From basic physics we know:

$t=\frac{d}{S1}$

so:

$t1=\frac{d}{S1}$

$t2=\frac{d}{S2}$

Using the previous information in equation (1)

$S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }$

Factoring:

$S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}$ (2)

Finally, replacing the data in (2)

$S_a_v_e_r_a_g_e=\frac{2*40*60}{60+40} =48km/h$

5 0

3 years ago

Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0

3 years ago

The right eye and right lung are __________.

PilotLPTM [1.2K]

The correct answer is A) Ipsilateral

Explanation:

The term ipsilateral is commonly used to describe objects or structures that are on the same side of a body or structure. This term is correct to describe the right eye and the right lung because these two organs are on the same side of the body (the right side). This can also be used to describe other organs such as the left humerus and the left hand or the right ear and the right feet because these pairs are also on the same side. According to this, the correct answer is A.

3 0

3 years ago

Where is Chemical Energy in a material stored ?

Sphinxa [80]

It is stored in the bonds between atoms

6 0

3 years ago