Answer:
20.08 Volts
Explanation:
Parallel Connection of Capacitors
The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is
[tex]\displaystyle V_1=\frac{Q_1}{C_1}[/tex]
[tex]\displaystyle V_2=\frac{Q_2}{C_2}[/tex]
They are both the same after connecting them, thus
[tex]\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}[/tex]
Or, equivalently
[tex]\displaystyle Q_2=\frac{C_2Q_1}{C_1}[/tex]
The total charge of both capacitors is
[tex]\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)[/tex]
We can compute the total charge by using the initial conditions where both capacitors were disconnected:
[tex]Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C[/tex]
Now we compute Q1 from the equation above
[tex]\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C[/tex]
The final voltage of any of the capacitors is
[tex]\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V[/tex]
