Respuesta :
- (a) [tex]Q = 1.70\times 10^{-2}\;\text{C}[/tex];
- (b) [tex]V_\text{final} = 5.31\times 10^{2}\;\text{V}[/tex];
- (c) [tex]E_\text{final} = 4.52\;\text{J}[/tex];
- (d) [tex]\Delta E = 2.82\;\text{J}[/tex].
All four values are in 3 sig. fig.
Explanation
(a)
[tex]Q = C\cdot V = 20.0\times 10^{-6} \times 850\;\text{V} = 1.70\times 10^{-2}\;\text{J}[/tex].
(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:
[tex]C_1\cdot V_\text{final} +C_2 \cdot V_\text{final} = C_1\cdot V_\text{initial}[/tex];
[tex](C_1+C_2)\cdot V_\text{final} = C_1\cdot V_\text{initial}[/tex];
[tex]\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}[/tex].
(c)
[tex]\displaystyle E = \frac{1}{2}\cdot C\cdot V^{2}[/tex].
[tex]\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}[/tex].
(d)
Initial energy of the system, which is the same as the initial energy in the [tex]20.0\;\mu\text{F}[/tex] capacitor:
[tex]\displaystyle E_\text{initial} = \frac{1}{2} \times 20.0\times 10^{-6} \times 850^{2} = 7.225\;\text{J}[/tex].
Change in energy:
[tex]\Delta E = 7.225\;\text{J} - 4.516\;\text{J} = 2.70\;\text{J}[/tex].
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 = [tex]Q_{1f} + Q_{2f}[/tex]
C₁ V₀ = C₁ [tex]V_{1f}[/tex] + C₂ [tex]V_{2f}[/tex]
the Power Difference final between the two capacitors must be the same, parallel connection
C₁ V₀ = (C₁ + C₂) [tex]V_f[/tex]
[tex]V_f[/tex] = [tex]\frac{C_1}{C_1+C_2} \ V_o[/tex]
V_f = [tex]\frac{20}{20+12} \ 850[/tex]
V_f = 531.25 V
c) The stored energy capacitor is
U = ½ C V²
The final energy system is
U = ½ (C₁ + C₂) [tex]V_f^2[/tex]
U = ½ (20 + 12) 10⁻⁶ 531.25²
U = 4.516 J
d) To calculate the energy change
ΔU = U₀ - [tex]U_f[/tex]
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
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