Answer to Question #312411 in Physics for keith

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

$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\frac{1}{C_4}$$ $$\frac{1}{C}=\frac{1}{2.5\:\rm \mu F}+\frac{1}{4.0\:\rm \mu F}+\frac{1}{6.0\:\rm \mu F}+\frac{1}{8.0\:\rm \mu F}$$ $$C=1.1\:\rm \mu F$$

The charge on the plates of each capacitor

$$q=CV\\ =1.1*10^{-6}*12.0=1.3*10^{-5}\:\rm C=13\:\mu C$$

The potential difference between plates of capacitors

$$V_1=\frac{q}{C_1}=\frac{13}{2.5}=5.2\:\rm V$$

$$V_2=\frac{q}{C_2}=\frac{13}{4.0}=3.2\:\rm V$$

$$V_3=\frac{q}{C_3}=\frac{13}{6.0}=2.1\:\rm V$$

$$V_4=\frac{q}{C_4}=\frac{13}{8.0}=1.6\:\rm V$$