In Fig. 25-40, C1 = 10.0 μF, 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?

Question

ganeshie8 · Answer

|dw:1451405945112:dw|

ganeshie8 · Answer

Here the answers :
 (a) 190 V; (b) 95 mJ

jskhupmang · Answer

you cheated yourself lmao

alexandervonhumboldt2 · Answer

stop spamming

irishboy123 · Answer

you know Q = CV and all that melarkey?

Q should be the same on each as the charge moves across and you add caps as you add parallel resistors

ganeshie8 · Answer

Ahh right, 

$v_i = \dfrac{q}{c_i} \le 100 \implies q \le 100c_i $ 

therefore maximum q is 100V*10 uF = 1 mC

irishboy123 · Answer

yes the 10 mic one is the one that will blow as $V = \frac{Q}{C}$ and Q will be same

irishboy123 · Answer

so if you have a 100 on the smallest, Q on each has to be 10 micF x 100V = 1 milli C

combined C i make as $\dfrac{1}{\dfrac{1}{10} + \dfrac{1}{20} +\dfrac{1}{25}} = \frac{100}{19} \mu C$

so $V_T = \frac{Q}{C_T} = \dfrac{10^{-3} \times 19}{100 \times 10^{-6}} = 190V$

checking on C2 we have $V_2 = \dfrac{10^{-3}}{20 . 10^-6 } = 50 $

checking on C3 we have $V_3 = \dfrac{10^{-3}}{25 . 10^-6 } = 40 $

irishboy123 · Answer

the energy will be easiest from something like $E_T = \frac{1}{2} C_{eq} V_T^2$

ganeshie8 · Answer

Wow this turned out to be simpler than I imagined !

$$\dfrac{1}{C_{eq}} = \dfrac{1}{10} + \dfrac{1}{20}+\dfrac{1}{25}$$
$$V_{\max} = \dfrac{q}{C_{eq}} =1mC*( \dfrac{1}{10} + \dfrac{1}{20}+\dfrac{1}{25})*10^6/F = 190V $$

ganeshie8 · Answer

$$E = \dfrac{1}{2}( \dfrac{1}{10} + \dfrac{1}{20}+\dfrac{1}{25})^{-1}*10^{-6}*190^2  = 95 mJ $$

Thank you @IrishBoy123  :)

irishboy123 · Answer

yes

it's very analogous to resistors, but keep your wits.  parallel resistor add up like series caps, and vice versa

the charge in series caps is the same as it is essentially a type of current [in the charge build up phase] , so the voltage drops are different.....as they are in series resistors

ganeshie8 · Answer

yeah $|q|$ on each capacitor plate has to be same in series combination

$V$ is ofcourse same across each capacitor in parallel combination