Question

15-mF capacitor and a 30-mf capacitor are connected in series, and charged to a potential difference of 50V. What is the resulting charge on the 30-mF capacitor?

Answer 1

Given:

• Capacotor 1, C1 = 15 mF = 15 x 10⁻³ F

,

• Capacitir 2, C2 = 30 mF = 30 x 10⁻³ F

,

• Potetntial difference, V = 50 V

Let's find the resulting charge on the 30-mF capacitor.

The charge on both capcitors will be the same since they are connected in series.

Given that they are connected in series, to find the resulting charge in the 30 mF capacitor, apply the formula:

$\frac{1}{C}=\frac{1}{15}+\frac{1}{30}$

Solving further:

$\begin{gathered} \frac{1}{C}=\frac{1}{15}+\frac{1}{30} \\ \\ \frac{1}{C}=\frac{2+1}{30} \\ \\ \frac{1}{C}=\frac{3}{30} \\ \\ \frac{1}{C}=\frac{1}{10} \\ \\ C=10\text{ mF = 10}\ast10^{-3}F \end{gathered}$

Now, apply the formula for charge:

$\begin{gathered} Q=CV \\ \\ Q=10\times10^{-3}\times50 \\ \\ Q=0.50mC \end{gathered}$

Therefore, the resulting charge on the 30-mF capacitor is 0.50 mC.

ANSWER:

0.50 mC.