Respuesta :
Answer:
A) 300 V
B) 44,447J
Explanation:
A) from the question, the capacitors are in series.
Therefore, the total potential between points A and B would be the sum total of potentials across each of the capacitors.
So, Maximum potential across A and B = [100 + 100 + 100]V = 300V
B) Let Cs be the effective capacitance of the series combination;
So, 1/Cs = (1/10) + (1/20) +(1/25)
1/Cs = 0.1 + 0.05 + 0.04
1/Cs = 0.19
Cs = 1/0.19 = 5.26 microF
Energy stored = (1/2) Cs(V)^(2)
= (1/2) x 5.26 x (130^(2)) = 44,447J
Part A. The maximum potential difference between points A and B is 300 V.
Part B. The maximum energy stored between points A and B is 26300 Joules.
How do you calculate the maximum potential difference and energy stored across the circuit?
Part A.
Given that the capacitors are connected in the series combination. The voltage across the circuit is 100V. So the total voltage is the sum of the voltage across each capacitor.
[tex]V = V_1+V_2+V_3[/tex]
[tex]V = 100 +100+100[/tex]
[tex]V = 300 \;\rm V[/tex]
Hence we can conclude that the maximum potential difference between points A and B is 300 V.
Part B.
The maximum capacitance across the circuit is calculated given below.
[tex]\dfrac {1}{C} = \dfrac {1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}[/tex]
[tex]\dfrac {1}{C} = \dfrac {1}{10}+\dfrac {1}{20}+\dfrac {1}{25}[/tex]
[tex]\dfrac {1}{C} = 0.1+0.05+0.04[/tex]
[tex]C = \dfrac{1}{0.19} = 5.26[/tex]
So the energy stored in the three capacitors is given below.
[tex]E = \dfrac{1}{2}CV^2[/tex]
[tex]E=\dfrac{1}2{}\times 5.26\times 100^2[/tex]
[tex]E = 26300 \;\rm J[/tex]
Hence we can conclude that the maximum energy stored between points A and B is 26300 Joules.
To know more about the voltage and energy, follow the link given below.
https://brainly.com/question/891089.
