Question

The capacitors have values C1 = 2.5 µF and C2 = 4.0 µF, C3 =

6.0 µF C4 = 8.0 µF and the potential difference across the battery is

12.0 V. Assume that the capacitors are connected in series. Find the

equivalent capacitance of the circuit and solve for the potential

difference across each capacitors.

Answer 1

Explanations & Calculations

• Capacitors add up their values to increase the net capacitance when connected parallelly while adding up to less when connected in series.
• The equivalent capacitance would be in this case,

$\qquad\qquad \begin{aligned} \small \frac{1}{C_{net}}&=\small \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}\\ \small C_{net}&=\small 1.06\\ &\approx \small 1.1\,\mu F \end{aligned}$

• To find the total charge that the entire circuit would hold, apply Q = CV formula,

$\qquad\qquad \begin{aligned} \small Q_{net}&=\small C_{net}.V\\ &=\small 1.06\,\mu F\times12V\\ &\approx\small 12.7\, \mu C \end{aligned}$

• Since all the capacitors are connected in series, each of them holds the same charge as the total charge. Then, the potential difference across each can be calculated using the same formula on each capacitor.

$\qquad\qquad \begin{aligned} \small V_{C1}&=\small Q/C_1\\ &=\small \frac{12.7\mu C}{2.5\mu F}\\ &\approx\small 5.1\, V \end{aligned}$

• Others can be evaluated similarly & you can try to get the answer.