Question

Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each other, positive terminal to positive terminal and negative terminal to negative terminal.(a) If C1 = 24 μF with an initial voltage of 25 V, and capacitor C2 = 13 μF is charged to 11 V. What is the final voltage, in volts, across C1 ?

Answer 1

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

$\displaystyle V_1=\frac{Q_1}{C_1}$

$\displaystyle V_2=\frac{Q_2}{C_2}$

They are both the same after connecting them, thus

$\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}$

Or, equivalently

$\displaystyle Q_2=\frac{C_2Q_1}{C_1}$

The total charge of both capacitors is

$\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)$

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

$Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C$

Now we compute Q1 from the equation above

$\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C$

The final voltage of any of the capacitors is

$\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V$