question_answer

A \[5\,\mu F\]capacitor is connected in series with a \[10\,\mu F\]capacitor. When a 300 V potential difference   is   applied   across   this combination/the total energy stored in the capacitors is

A) 15 J   

B) 1.5 J  

C)  0.15 J 

D)  0.10 J

Correct Answer: C

Solution :

 According to question, the figure can be drawn as below The equivalent capacitance, \[\frac{1}{{{C}_{eq}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}=\frac{1}{5}+\frac{1}{10}=\frac{2+1}{10}=\frac{3}{10}\] \[\Rightarrow \] \[{{C}_{eq}}=\frac{10}{3}\mu F\] Now, the energy stored in the capacitor is \[U=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\times \frac{10}{3}\times {{10}^{-6}}\times 300\times 300\] \[=\frac{3}{10\times 2}=\frac{3}{20}=0.15\,J\]