Respuesta :
Answer:
Explanation:
Given that,
First Capacitor is 10 µF
C_1 = 10 µF
Potential difference is
V_1 = 10 V.
The charge on the plate is
q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC
q_1 = 100 µC
A second capacitor is 5 µF
C_2 = 5 µF
Potential difference is
V_2 = 5V.
Then, the charge on the capacitor 2 is.
q_2 = C_2 × V_2
q_2 = 5µF × 5 = 25 µC
Then, the average capacitance is
q = (q_1 + q_2) / 2
q = (25 + 100) / 2
q = 62.5µC
B. The two capacitor are connected together, then the equivalent capacitance is
Ceq = C_1 + C_2.
Ceq = 10 µF + 5 µF.
Ceq = 15 µF.
The average voltage is
V = (V_1 + V_2) / 2
V = (10 + 5)/2
V = 15 / 2 = 7.5V
Energy dissipated is
U = ½Ceq•V²
U = ½ × 15 × 10^-6 × 7.5²
U = 4.22 × 10^-4 J
U = 422 × 10^-6
U = 422 µJ
The equivalent capacitance is 15μC and the energy dissipated is 422μJ.
Equivalent capacitance and energy dissipated:
The 10 µF capacitor has a potential difference of 10 V.
[tex]C = 10 \mu F\\\\V = 10 V[/tex]
The charge on the plate is given by:
[tex]Q = CV \\\\Q = 10 \times10^{-6} \times 10[/tex]
Q = 100µC
The 5 µF capacitor has a potential difference of 5 V.
[tex]C' = 10 \mu F\\\\V' = 10 V[/tex]
The charge on the plate is given by:
[tex]q = C'V' \\\\q = 5 \times10^{-6} \times 5[/tex]
q = 25µC
The average capacitance is given by:
Q(avg) = (Q + q) / 2
Q(avg) = (25 + 100) / 2
Q(avg) = 62.5µC
The equivalent capacitance for capacitors connected in parallel is given by:
[tex]C_{eq} = C + C'\\\\C_{eq} = 10 \mu F + 5 \mu F\\\\C_{eq} = 15 \mu F[/tex]
The average voltage is:
V = (V + V') / 2
V = (10 + 5)/2
V = 7.5 V
The energy dissipated is:
[tex]U = \frac{1}{2}C_{eq}V^2\\\\U = 0.5 \times 15 \times 10^{-6} \times 7.5^2\\\\U = 422 \times 10^{-6}J[/tex]
U = 422 µJ
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