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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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A Christmas light is made to flash via the discharge of a capacitor. The effective duration of the flash is 0.25 s (which you ca

Sonbull [250]

Answer:

The correct solution is:

(a) $1.375\times 10^{-2} \ J$

(b) $4.43\times 10^{-3} \ C$

(c) $1.42\times 10^{-3} \ F$

(d) $178.57 \ \Omega$

Explanation:

The given values are:

Effective duration of the flash,

ζ = 0.25 s

Average power,

$P_{avg}=55 \ mW$

$=55\times 10^{-3} \ W$

Average voltage,

$V_{avg}=3.1 \ V$

Now,

(a)

⇒ $E=P_{avg}\times \zeta$

On substituting the values, we get

⇒ $=55\times 10^{-3}\times 0.25$

⇒ $=1.375\times 10^{-2} \ J$

(b)

⇒ $E=Q\times V_{avg}$

then,

⇒ $Q=\frac{E}{V_{avg}}$

On substituting the values, we get

⇒ $=\frac{1.375\times 10^{-2}}{3.1}$

⇒ $=4.43\times 10^{-3} \ C$

(c)

⇒ $C=\frac{Q}{V}$

⇒ $=\frac{4.43\times 10^{-3}}{3.1}$

⇒ $=1.42\times 10^{-3} \ F$

(d)

As we know,

⇒ $R=\frac{1}{4C}$

⇒ $=\frac{1}{4\times 1.42\times 10^{-3}}$

⇒ $=\frac{1000}{5.6}$

⇒ $=178.57 \ \Omega$

5 0

2 years ago

What is meant by the "lever arm" of a torque?

enyata [817]

Torque = Force applied x lever arm. It twists the object or changes the rotational motion of things

Hope it helps

3 0

3 years ago

3. Given f(x) = x2 - 5x+3, then A. f(-1) = -3 b. f(-1) = 7 f(-1) = -1 d. f(-1) = 9

Phantasy [73]

Answer:

$f(x) = {x}^{2} - 5x + 3 \\ f( - 1) = {( - 1)}^{2} - 5( - 1) + 3 \\ = 9$

7 0

3 years ago

If the velocity of a pitched ball has a magnitude of 44.5 m/s and the batted ball's velocity is 55.5 m/s in the opposite directi

Lelu [443]

Answer:

ΔP = 14.5 Ns

I = 14.5 Ns

ΔF = 5.8 x 10³ N = 5.8 KN

Explanation:

The mass of the ball is given as 0.145 kg in the complete question. So, the change in momentum will be:

ΔP = mv₂ - mv₁

ΔP = m(v₂ - v₁)

where,

ΔP = Change in Momentum = ?

m = mass of ball = 0.145 kg

v₂ = velocity of batted ball = 55.5 m/s

v₁ = velocity of pitched ball = - 44.5 m/s (due to opposite direction)

Therefore,

ΔP = (0.145 kg)(55.5 m/s + 44.5 m/s)

ΔP = 14.5 Ns

The impulse applied to a body is equal to the change in its momentum. Therefore,

Impulse = I = ΔP

I = 14.5 Ns

the average force can be found as:

I = ΔF*t

ΔF = I/t

where,

ΔF = Average Force = ?

t = time of contact = 2.5 ms = 2.5 x 10⁻³ s

Therefore,

ΔF = 14.5 N.s/(2.5 x 10⁻³ s)

ΔF = 5.8 x 10³ N = 5.8 KN

4 0

3 years ago

Please help I'm stuck on this question

Afina-wow [57]

Answer:

increase

decrease

Explanation:

using formula

Vt=mg/6πηr

so if m increases V increases

r is the denominator so if r increases V decreases

8 0

2 years ago