Question

A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is?

0.05 C and 5 J

0.08 C and 12 J

0.05 C and 10 J

0.02 C and 4 J​

Answer 1

Answer:-

• Capacitance=C=50µf
• potential Difference=400V
• Charge=Q=?

We know that

$\boxed{\sf Q=CV}$

$\\ \qquad\quad\sf\rightarrowtail Q=50\times 400$

$\\ \qquad\quad\sf\rightarrowtail Q=20000$

$\\ \qquad\quad\sf\rightarrowtail Q=2\times 10^4C$

And

$\boxed{\sf E_{cap}=\dfrac{CV^2}{2}}$

$\\ \qquad\quad\sf\rightarrowtail E_{cap}=\dfrac{50\times \cancel{(400)}^2}{\cancel{2}}$

$\\ \qquad\quad\sf\rightarrowtail E_{cap}=\dfrac{50\times (200)^2}{1}$

$\\ \qquad\quad\sf\rightarrowtail E_{cap}=50\times 40000$

$\\ \qquad\quad\sf\rightarrowtail E_{cap}=2000000J$

$\\ \qquad\quad\sf\rightarrowtail E_{cap}=2\times 10^6J$

$\\ \qquad\quad\sf\rightarrowtail E_{cap}=2\times 10^3KJ$