Question

A 1.50-V battery is used to charge a capacitor. When the capacitor is fully charged, it stores 64.0 mJ. If the capacitor were fully charged by a 3.00-V battery instead, how much energy would it store?

Answer 1

Answer:

256.05 mJ

Explanation:

The energy (E) stored in a capacitor when it is charged by a battery of voltage,V, is related to the capacitance, C, of the capacitor as follows;

E  = $\frac{1}{2}$ x C x V²         ---------------(i)

From the question;

When a 1.50V battery is used, the energy is 64.0mJ. This means that when;

V = 1.50V, E = 64.0mJ = 0.064 J

Substitute these values into equation (i) to find the capacitance of the capacitor as follows;

0.064 = $\frac{1}{2}$ x C x 1.5²

=> 0.064 = $\frac{1}{2}$ x C x 2.25

=> 0.064 = 1.125 x C

=> C = $\frac{0.064}{1.125}$

Solve for C;

=> C = 0.0569F

Now, the we know the capacitance of the capacitor, let's calculate how much energy will be stored if the battery were 3.00V by substituting V = 3.00V and C = 0.0569 into equation (i) as follows;

E = $\frac{1}{2}$ x 0.0569 x 3.00²  

E = $\frac{1}{2}$ x 0.0569 x 9.00

E = 0.25605 J

Multiply the result by 1000 to convert it to mJ. i.e;

E = (0.25605 x 1000) mJ

E = 256.05 mJ

Therefore, the amount of energy it would store if the capacitor were fully charged by a 3.00V battery is 256.05 mJ

Answer 2

Answer:

0.256 J.

Explanation:

The formula for the energy stored in a capacitor is given as

E = 1/2CV².................... Equation 1

Where E = Energy stored in the capacitor, C = Capacitance of the capacitor, V = Voltage.

Make C the subject of the equation

C = 2E/V².................. Equation 2

Given: E = 64 mJ  = 0.064 J, V = 1.5 V.

Substitute into equation 2

C = 2(0.064)/(1.5²)

C = 0.128/2.25

C = 0.0569 F.

If the capacitor where fully charged by 3.00 V.

E = 1/2CV²

Given: C = 0.0569 F, V = 3.0 V

Substitute into the equation above,

E = 1/2(0.0569)(3²)

E = 0.256 J.