Electrostatic potential and capacitance

[vipulgoyal]

Homework Statement

C₁20 uF C₂=40uF C₃= 50uF if no capacitor has the capacity to bear more than 50V then maximum potential difference between the two ends is

Homework Equations

Q=CV

The Attempt at a Solution

i tried by calculating the C_eq then putting it in the equation finding out q i know it doesn't make sense...

Just help me..

 

[cupid.callin]

let emf of battery is V

Find net q through battery

then find q on each capacitor and thus calculate voltage of each

Now maximum voltage you find cannot exceed

then calculate V

 

[vipulgoyal]

let emf of battery is V

Find net q through battery

then find q on each capacitor and thus calculate voltage of each

Now maximum voltage you find cannot exceed

then calculate V

q(battery) = q on each capacitor as the conection is made in series (sorry forgot to mention that
)

 

[vipulgoyal]

nyways answer will not come by this method it will result into the given variable

let V be the emf of battery
then q= C (eq) V
q= 200/19 V

now you suggest to find V' thre are two variables above so can't find that either

 

[cupid.callin]

q = 200/19 V

so potential on each capacitor is : (200V/19)/20 , (200V/19)/40 , (200V/19)/50

now the maximum of these is equal to 50 (given)

so equate and find V

answer:

Spoiler

V = 95