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 \[U=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\times \frac{10}{3}\times {{10}^{-6}}\times 300\times 300\] \[=\frac{3}{10\times 2}=\frac{3}{20}=0.15\,J\]