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2 years ago

C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F. Calculate the equivalent capacitance

Physics

1 answer:

grin007 [14]2 years ago

6 0

The equivalent capacitance between A and B points is 2.5F.

<h3>What is parallel plate capacitor?</h3>

The two parallel plates placed at a distance apart used to store charge when electric supply is on.

The capacitance of a capacitor is given by

C = ε₀ A/d

From the given circuit C1, C2 and C3, C4 are in parallel C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F

C1, C2 = 4 +4 =8F

C3, C4 = 2 +4 =6F

Now , all capacitors are in series.

Total equivalent capacitance is
1 / Ceq = 1/ 8 +1/6 +1/ 9.2

Ceq = 2.5 F

Thus, the equivalent capacitance between A and B points is 2.5F.

Learn more about parallel plate capacitor.

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3 years ago

Read 2 more answers

A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch

ANTONII [103]

Answer:

A. A = 0.913 m

B. amax = 132.24m/s^2

C. Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

$v_{max}=A\omega$ (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

$\omega=\sqrt{\frac{k}{m}}$ (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

$\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}$

Next, you solve the equation (1) for A and replace the values of vmax and w:

$A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m$

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

$a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}$

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

$F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N$

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

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3 years ago

An external resistor with resistance R is connected to a battery that has an emf E and an internal resistance r. Let P be the el

satela [25.4K]

Answer:

a) When R is very small R << r, therefore the term R+ r will equal r and the current becomes

b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes

Explanation:

<u>Solution :</u>

(a) We want to get the consumed power P when R is very small. The resistor in the circuit consumed the power from this battery. In this case, the current I is leaving the source at the higher-potential terminal and the energy is being delivered to the external circuit where the rate (power) of this transfer is given by equation in the next form

P=∈*I-I^2*r (1)

Where the term ∈*I is the rate at which work is done by the battery and the term I^2*r is the rate at which electrical energy is dissipated in the internal resistance of the battery. The current in the circuit depends on the internal resistance r and we can apply equation to get the current by

I=∈/R+r (2)

When R is very small R << r, therefore the term R+ r will equal r and the current becomes

I= ∈/r

Now let us plug this expression of I into equation (1) to get the consumed power

P=∈*I-I^2*r

=I(∈-I*r)

The consumed power when R is very small is zero

(b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes

I=∈/R

The dissipated power due toll could be calculated by using equation.

P=I^2*r (3)

Now let us plug the expression of I into equation (3) to get P

P=I^2*R=(∈/R)^2*R

=∈^2/R

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3 years ago