Physics, 03.03.2020 02:57, arroyoe342

C2 = 20.0 μF, and C3 = 25.0 μF. If no capacitor can withstand a potential difference of more than 100 V without failure, what are (a) the magnitude of the maximum potential difference that can exist between points A and B and (b) the maximum energy that can be stored in the three-capacitor arrangement?

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answered: Tabbicat021

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.

V = V_1+V_2+V_3

V = 100 +100+100

$V = 300 \;\rm V$

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.

$\dfrac {1}{C} = \dfrac {1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}$

$\dfrac {1}{C} = \dfrac {1}{10}+\dfrac {1}{20}+\dfrac {1}{25}$

$\dfrac {1}{C} = 0.1+0.05+0.04$

$C = \dfrac{1}{0.19} = 5.26$

So the energy stored in the three capacitors is given below.

$E = \dfrac{1}{2}CV^2$

$E=\dfrac{1}2{}\times 5.26\times 100^2$

$E = 26300 \;\rm J$

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.

link.

answered: clietmaster112

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

answered: Guest

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.

answered: Guest

work: w = fd (given)

work done by chet =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

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C2 = 20.0 μF, and C3 = 25.0 μF. If no capacitor can withstand a potential difference of more than 10...