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You have two different batteries, 6 V and 12 V, and two different capacitors, 0.1 F and 1.0 F. Which combination of battery and capacitor will store the most energy in the capacitor?

A. 6 V battery (either capacitor)
B. 6 V & 1.0 F
C. 12 V battery (either capacitor)
D. 12 V & 0.1 F
E. 6 V & 0.1 F
F. 1.0 F capacitor (either battery)
G. 12 V & 1.0 F
H. 0.1 F capacitor (either battery)

Answer :

The combination of battery and capacitor that will store the most energy in the capacitor is option g. 12 V & 1.0 F.

To determine this, we need to calculate the energy stored in each combination using the formula for the energy stored in a capacitor:

Energy = (1/2) × C × V²

Where:
- Energy is the energy stored in the capacitor (in Joules)
- C is the capacitance of the capacitor (in Farads)
- V is the voltage of the battery (in Volts)

Using this formula, we can calculate the energy stored in each combination and find the one with the highest value. Option g (12 V & 1.0 F) will have the most energy stored:

Energy = (1/2) × 1.0 F × (12V² = 72 Joules

You can learn more about capacitors at: brainly.com/question/29301875

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Rewritten by : Barada

The 12 V & 1.0 F capacitor combination stores the most energy, with an energy storage of 72 joules. Hence option g is the answer.

The combination of battery and capacitor that will store the most energy in the capacitor is determined by the energy storage formula for a capacitor, which is [tex]Energy = \frac{1}{2} X Capacitance X (Voltage)^2[/tex]. For the given options:

6V & 0.1 F:

[tex]Energy = \frac{1}{2} X 0.1 \ F X (6 \ V)^2\\ Energy = 1.8 \ J[/tex]

6V & 1.0 F:

[tex]Energy = \frac{1}{2} X 1.0 \ F X (6 \ V)^2 \\E = 18 \ J,[/tex]

12 V & 0.1 F:

[tex]Energy = \frac{1}{2} X 0.1 \ F X (12 \ V)^2 \\E = 7.2 \ J,[/tex]

12 V & 1.0 F:

[tex]Energy = \frac{1}{2} X 1.0\ F X (12 \ V)^2 \\E = 72 \ J[/tex]

Therefore, the combination that stores the most energy is the 12 V & 1.0 F capacitor, with an energy storage of 72 joules.