The three capacitors in figure, store a total energy in µJ of
Answer & explanation
Correct answer: option 3
Total capacitance = 6 µF
Applied voltage = 4v
Stored energy U = $\frac{1}{2} C V^2=\frac{1}{2} \times 6 \times 10^{-6} \times(4)^2$ = 48µF
The three capacitors in figure, store a total energy in µJ of
Correct answer: option 3
Total capacitance = 6 µF
Applied voltage = 4v
Stored energy U = $\frac{1}{2} C V^2=\frac{1}{2} \times 6 \times 10^{-6} \times(4)^2$ = 48µF