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2 years ago

What would happen when two or more capacitors are connected in parallel across a potential difference?

Physics

1 answer:

Varvara68 [4.7K]2 years ago

6 0

Answer:

The Current Iₜ = I₁ + I₂ + I₃

Charge Qₜ = Q₁ + Q₂ + Q₃

Potential difference Vₜ = V₁ = V₂ = V₃

The total capacitance Cₜ = C₁ + C₂ + C₃

Explanation:

According to the attached image;

For parallel arrangements of capacitors, the current flowing through each of the capacitors sums up to the total current flowing through the circuit;

Iₜ = I₁ + I₂ + I₃

Also the charge storage by each capacitor sums up to give the total charge stored;

Qₜ = Q₁ + Q₂ + Q₃

The potential difference across each of the capacitors are the same and equal to the total voltage across the circuit;

Vₜ = V₁ = V₂ = V₃

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

Cₜ = C₁ + C₂ + C₃

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If 250 g of calcium chlorate decomposes into calcium chloride and oxygen, what mass of oxygen would be produced?

sineoko [7]

115.94g of O₂ are produced by 250g of Ca(ClO₃)₂

<u>Explanation:</u>

Ca(ClO₃)₂ -->CaCl₂ + 3O₂

One mole of calcium chlorate give three moles of oxygen.

Molar mass of calcium chlorate is

40+(35.5+16*3)*2 = 207g/mol

Molar mass of Oxygen = 32g

1 mole of calcium chlorate decomposes to form 3 moles of oxygen.

So,

207g of Ca(ClO₃)₂ produces 3 X 32g of O₂

250g of Ca(ClO₃)₂ will produce = $\frac{3 X 32}{207} X 250$

= 115.94g of O₂

Therefore, 115.94g of O₂ are produced by 250g of Ca(ClO₃)₂

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2 years ago

/How much kinetic energy does an 80 kg man have while running at 1.5 m/s

igomit [66]

K.E = 0.5 * m * v^2
= 0.5 * 80 * (1.5)^2
K.E = 90 joule

7 0

2 years ago

two technicians are discussing fuses technician a says that current flows should be 20% less than fuse rating technician B says

MrMuchimi

The technician A is correct. The current should be 20% or any given percentage lower than the current for which the fuse is created.

The technician B argues that orange blade fuses have a 9 Amp rating. But they usually have a 40A rating

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2 years ago

A certain elevator cab has a total run of 198 m and a maximum speed is 317 m/min, and it accelerates from rest and then back to

tester [92]

Answer:

Explanation:

Given

Total run=198 m

maximum speed=317 m/min=5.28 m/s

acceleration $=1.11 m/s^2$

(a)Distance traveled while accelerating

$v^2-u^2=2as$

$(5.28)^2-0=2\times 1.11\times s$

s=12.57 m

time taken

v=u+at

$5.28=0+1.11\times t$

t=4.75 s

(b)Time taken to cover 198 m

if car takes 12.57 m & 4.75 s to reach max speed so it also takes 12.57 m and 4.75 s to stop from max speed

Distance traveled with max speed=198-25.14=172.86 m

time taken $=\frac{172.86}{5.28}=32.73 s$

total time=4.75+32.73+4.75=42.25 s

7 0

2 years ago

An astronaut is on a 100-m lifeline outside a spaceship, circling the ship with an angular speed of

nataly862011 [7]

Answer:

D = 72.68 m

Explanation:

given,

R = 100 m

angular speed = 0.1 rad/s

distance she can be pulled before the centripetal acceleration reaches 5g = 49 m/s².

using conservation of Angular momentum

$I_i\omega_i= I_f\omega_f$

$mr_i^2\omega_i=m r_f^2\omega_f$

$\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i$

we know,

centripetal acceleration

$a = \dfrac{v^2}{r}$

v = r ω

$a =\omega_f^2 r_f$

$a =(\dfrac{r_i^2}{r_f^2}\times \omega_i)^2 r_f$

$a =\dfrac{r_i^4\times \omega_i^2}{r_f^3}$

$r_f^3=\dfrac{100^4\times 0.1^2}{5\times 9.8}$

$r_f^3=20408.1632$

$r_f = 27.32\ m$

distance she has reached inward is equal to

D = 100 - 27.23

D = 72.68 m

3 0

2 years ago