Answer to Question #311529 in Electricity and Magnetism for Maymay

Question

Three capacitors with capacitances 4 F, and 12 F are connected in series. A voltage of 400V is applied to the combination.Find the charge in each capacitor.(IN C)

Answer 1

When capacitors was combined in series, charge on each capacitor is equal total charge. Find total capacity in series

$$C=\frac{C_1C_2}{C_1+C_2}$$

For total charge

$$Q=CU$$

Answer 1200 C

Calculating

$$Q=3\cdot400=1200$$ $$C=\frac{4\cdot12}{4+12}=3$$