Question

10. The energy stored in 4 uF capacitors is??​

Answer 1

Answer:

use E =1/2 CV^2

add capacitances and take equivalent capacitance

Answer 2

Answer:

Energy stored in the capacitor.

Explanation:

Concepts:

Capacitor in series: When two or more capacitors are in series we use

$\frac{1}{C_{s}} =\frac{1}{C_{1} } +\frac{1}{C_{2} } +.....$

Capacitor in parallel: When two or more capacitors are in parallel we use;

${C_{p} }={C_{1} }+{C_{2} } ......$

Energy of capacitor: The energy of capacitor is written as;

$E=\frac{1}{2} CV^{2}$

Calculations:

Given:

The in given diagram we have three capacitors.

Let $C_{1} = 2$μF

$C_{2} = 6$μF

$C_{3}=4$μF

V = 12 V

Step 1:

Here $C_{2},C_{3}$ are parallel to each other therefore we have;

${C_{p} ={C_{2} } +{C_{3} }$

$C_{p}=6+4$

    = 10μF

Now $C_{p}$ and $C_{1}$ is in the series therefore we get;

$\frac{1}{C_{s'} } =\frac{1}{10} +\frac{1}{2} \\\frac{1}{C_{s'} } =\frac{1+5}{10} \\\frac{1}{C_{s'} } =\frac{3}{5} \\C_{s'} =\frac{5}{3}$

Step 2:

Now we know that

Q = CV

Q= $\frac{5}{3}$×$12$

Q=20

Voltage difference between $C_{p}$ and $C_{1}$ is written as:

$V=12-\frac{20}{2} \\V= 2V$

Step 3:

Therefore,

Energy ,E=$\frac{1}{2}$×$4$×$2^{2}$

E= 8μJ

Conclusion:

The energy stored is equal to 8μJ. Option (1) is correct.