Question

Two capacitors of capacities \(5 \, \mu{F} \) and \( 10 \, \mu{F} \)F respectively are connected in series. Calculate the resultant capacity of the combination.

Hint:

When capacitors are connected in series, the equivalent capacitance is always smaller than the smallest individual capacitance.

Answer 1

For capacitors connected in series, the equivalent capacitance \( C_{{eq}} \) is given by the formula: \[ \frac{1}{C_{{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \] Substituting \( C_1 = 5 \, \mu{F} \) and \( C_2 = 10 \, \mu{F} \): \[ \frac{1}{C_{{eq}}} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10} \] \[ C_{{eq}} = \frac{10}{3} = 3.33 \, \mu{F} \] \bigskip Next, using the formula for the energy stored in a capacitor: \[ E = \frac{1}{2} C_{{eq}} V^2 \] where \( V \) is the voltage across the capacitor combination, we can calculate the energy. For \( V = 10^2 \, {V} \), the energy becomes: \[ E = \frac{1}{2} \times 3.33 \times 10^{-6} \times (100)^2 = 2.4 \times 10^{-19} \, {J}. \] \bigskip