Can someone please tell me why I\'m wrong :-) thanks! Three 0.16 AuF capacitors
ID: 2000380 • Letter: C
Question
Can someone please tell me why I'm wrong :-) thanks!
Three 0.16 AuF capacitors are connected in parallel across a 12 V battery, as shown in the figure below. The battery is then disconnected. Next, one capacitor is carefully disconnected so that it doesn't lose any charge and is reconnected backward, that is, with its positively charged side and its negatively charged side reversed 12 V (a) What is the potential difference across the capacitors now? 12 X V (b) By how much has the stored energy of the capacitors changed in the process? Three capacitors (C o.16 x 10-6 F) are connected in parallel across a battery (Vi 12 v). We can use q CV to calculate the initial charge on each capacitor. The battery is disconnected and then the middle capacitor is quickly disconnected and reconnected backward. Since we've reversed one of the capacitors, the total net charge in the circuit will decrease and redistribute equally amongst the three capacitors. After calculating the final charge on each capacitor we can CV again to find the final potential difference across the capacitors. The energy stored in each capacitor is given by use q CV We can set up a ratio to calculate by how much the energy stored in each capacitor changes after reversing the middle capacitor.Explanation / Answer
a) when battery is disconnected , charge will remain same. It can be calculated as
q1=c1 * v
q1= 0.16*10^-6 * 12
= 1.92 * 10^-6 coulomb
it will remain constant . but if battery remains disconnected then potential difference will reduce to zero because charge remains constant when battery is disconnected not potential difference.
"I think you recalculated voltage using the same formula q=cv where q and c are known to us"
b) q1= q2= q3 =1.92 *10^-6 coulomb
total charge = q= q1+q2+q3
energy stored = u = q^2 / (2C)
u(final) = u (initial) as total charge 'q' remains same after the battery gets disconnected and total capcaitance also remains same as they will be in parallel combination
So u(final)/u(initial) = 1