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3 years ago

What would happen when two or more capacitors are connected in series across a potential difference, then

Physics

2 answers:

anyanavicka [17]3 years ago

6 0

Answer:

The Current Iₜ = I₁ = I₂ = I₃

Charge Qₜ = Q₁ = Q₂ = Q₃

Potential difference Vₜ = V₁ + V₂ + V₃

Capacitance 1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃

Explanation:

According to the attached image;

For series arrangements of capacitors, the current flowing through each of the capacitors are the same and equal to the total current flowing through the circuit;

Iₜ = I₁ = I₂ = I₃

Also the charge storage by each capacitor is equal;

Qₜ = Q₁ = Q₂ = Q₃

The potential difference across each of the capacitors sum up to the total voltage across the circuit;

Vₜ = V₁ + V₂ + V₃

The reciprocal of the total capacitance equals the sum of the reciprocal of capacitances of each of the capacitors;

1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃

crimeas [40]3 years ago

6 0

Answer:

1) the net potential difference of the circuit will be the sum of all the potential difference of all the individual capacitors that are connected together.

2) The net capacitance in the circuit would be lesser than that of the capacitance of the individual capacitors.

3) the charges on the individual capacitors will remain the same.

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You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp

GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(d) $539.55\mu$

(e) 0

Explanation:

The period for 1 circle $2\pi$ of the merry go around is 9.5s. It means the angular speed is:

$\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s$

(a)The speed is

$v = \omega * R = 0.661 * 17 = 11.24 m/s$

(b) Centripetal acceleration:

$a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2$

$F = ma = 55 * 7.44 = 409 N$

(d) let the coefficient of friction by $\mu$ . The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

$F_f = mg\mu = 55*9.81\mu = 539.55\mu$

3 0

3 years ago

Two blocks with different temperatures had entropies of 10 J/K and 30 J/K before they were brought in contact. What can you say

True [87]

Answer:

a. Ssystem > 40 J/K

Explanation:

Given that

The entropy of first block = 10 J/K

The entropy of second block = 30 J/K

When two bodies come into contact with each other, the entropy of the combined system will increase and the entropy sum remains unchanged: According to the Second law of thermodynamics.The entropy of the system will be greater than 40 J/K.

Therefore the answer is a.

Ssystem > 40 J/K

6 0

3 years ago

(TCO 5) The relationship between Celsius (º C) and Fahrenheit (º F) degree of measuring temperature is linear. Find the linear e

gtnhenbr [62]

Answer:

$C=-\dfrac{10}{17}(F-32)$