Physics, 30.09.2019 00:30 aiken11192006

A20.0 μf capacitor is charged to a potential difference of 850 v. the terminals of the charged capacitor are then connected to those of an uncharged 12.0 μf capacitor. (a) compute the original charge of the system. (b) compute the final potential difference across capacitor. (c) compute the final energy of the system. (d) compute the decrease in energy when the capacitors are connected.

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Answer from: taylorray0820

The properties of the capacitors can be calculated the answers are:

a) Q₁ = 1.70 10-2 C

b) V_f = 531 V

c) U_f = 4.52 J

d) ΔU = 2.71 J

Given parameters

The capacitance C₁ = 20.0 10⁻⁶ F and C₂ = 12.0 10⁻⁶ F The initial potential deference V1 = 850 V

To find

a) The initial charge

b) The potential difference of the system connected capacitors

c) The final energy of the system

d) the energy change when connecting the capacitors

A capacitor is a system formed by two separate parallel plates that serves to store electrical charge,

Q = C V

Where Q is the stored charge, C the capacitance and V the potential difference

a) ask for the initial charge

Q₀ = C₁ V₀

Q₀ = 20.0 10⁻⁶ 850

Q₀ = 1.70 10⁻² C

b) The law of conservation of charge establishes that the electric charge cannot be created or destroyed, therefore the initial charge (Q₀) must be distributed between the two connected capacitors

Q₀ = $Q_{1f} + Q_{2f}$

C₁ V₀ = C₁ $V_{1f}$ + C₂ $V_{2f}$

the Power Difference final between the two capacitors must be the same, parallel connection

C₁ V₀ = (C₁ + C₂) V_f

V_f = $\frac{C_1}{C_1+C_2} \ V_o$

V_f = $\frac{20}{20+12} \ 850$

V_f = 531.25 V

c) The stored energy capacitor is

U = ½ C V²

The final energy system is

U = ½ (C₁ + C₂) V_f^2

U = ½ (20 + 12) 10⁻⁶ 531.25²

U = 4.516 J

d) To calculate the energy change

ΔU = U₀ - U_f

let's look for the initial energy

U₀ = ½ C₁ V₀²

U₀ = ½ 20 10⁻⁶ 850²

U₀ = 7.225 J

whereby the energy change is

ΔU = 7.225 - 4.516

ΔU = 2.71 J

In conclusion using the properties of the capacitors we were able to calculate the answers are:

a) Q₁ = 1.70 10-2 C

b) V_f = 531 V

c) U_f = 4.52 J

d) ΔU = 2.71 J

Learn more about capacitors here:

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Answer from: MIC620

(a) $Q = 1.70\times 10^{-2}\;\text{C}$ ;(b) $V_\text{final} = 5.31\times 10^{2}\;\text{V}$ ;(c) $E_\text{final} = 4.52\;\text{J}$ ;(d) $\Delta E = 2.82\;\text{J}$ .

All four values are in 3 sig. fig.

Explanation

(a)

$Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}$ .

(b)

Sum of the final charge on the two capacitors should be the same as the sum of the initial charge. Voltage of the two capacitors should be the same. That is:

$C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$(C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial}$ ;

$\displaystyle V_\text{final} = \frac{C_1}{C_1+C_2}\cdot V_\text{initial}\\\phantom{V_\text{final}} = \frac{20.0\;\mu\text{F}}{20.0\;\mu\text{F} + 12.0\;\mu\text{F}} \times 850\;\text{V}\\\phantom{V_\text{final}} =531\;\text{V}$ .

(c)

$\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}$ .

$\displaystyle E_\text{final} = \frac{1}{2} (C_1 + C_2) \cdot {V_\text{final}}^{2} \\\phantom{E_\text{final}} = \frac{1}{2} \times (20.0\times 10^{-6} + 12.0\times 10^{-6}) \times 531.25\\\phantom{E_\text{final}} = 4.52\;\text{J}$ .

(d)

Initial energy of the system, which is the same as the initial energy in the $20.0\;\mu\text{F}$ capacitor:

$\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}$ .

Change in energy:

$\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}$ .

Answer from: Quest

i don't understand. explain it clearly

Answer from: Quest

a=d/t

a="triangle"v/t

explanation:

a = acceleration

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