Two capacitors of capacitances \( 20\,\mu\mathrm{F} \) and \( 60\,\mu\mathrm{F} \) are connected in series in an electric circuit with a battery of \( 20\,\mathrm{V} \). What will be the charge on each capacitor?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 300\,\mu\mathrm{C} \)
In a series combination every capacitor carries the same charge, equal to the charge drawn from the battery by the equivalent capacitance.
\( \displaystyle \frac{1}{C_{s}} = \frac{1}{20} + \frac{1}{60} = \frac{3+1}{60} = \frac{1}{15} \)
\( C_{s} = 15\,\mu\mathrm{F} \)
\( Q = C_{s}V = (15\times10^{-6}\,\mathrm{F})(20\,\mathrm{V}) = 300\times10^{-6}\,\mathrm{C} = 300\,\mu\mathrm{C} \)
Check: \( V_{1} = 300/20 = 15\,\mathrm{V} \) and \( V_{2} = 300/60 = 5\,\mathrm{V} \), which add up to 20 V.