Question

A $6\mu F$ capacitor is charged from $10V$ to $20V$. An increase in energy will be?

Answer 1

Step 1: Given Data

The capacitance of the capacitor $C=6\mu F=6\times {10}^{-6}F$

Initial voltage $V_i=10V$

Final voltage $V_f=20V$

Step 2: Formula Used

The energy of a capacitor is given as

$E=\frac{1}{2}C{V}^2$

Therefore the change in energy can be given as

$∆E=\frac{1}{2}C{v_f}^2-\frac{1}{2}C{v_i}^2$

$∆E=\frac{1}{2}C\left({v_f}^2-{v_i}^2\right)$

Step 3: Calculate the increase in energy

Upon substituting the values we get,

$∆E=\frac{1}{2}\times 6\times {10}^{-6}\left({20}^2-{10}^2\right)$

$=3\times {10}^{-6}\left(20+10\right)\left(20-10\right)$

$=3\times {10}^{-6}\left(30\right)\left(10\right)$

$=9\times {10}^{-4}J$

Hence, the increase in energy of the capacitor will be $9\times {10}^{-4}J$.