You have two capacitors and want to connect them across a voltage source (battery) to store the maximum amount of energy. Should they be connected in series or in parallel?
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You have two capacitors and want to connect them across
a voltage source (battery) to store the maximum amount of energy.
Should they be connected in series or in parallel?
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- In the figure how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which κ = 2.10; both capacitors have a plate area of 4.70 × 10–3 m2 and a plate separation of 6.00 mm.(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 24.0 µC. What is the voltage of the battery? V (b) If the same capacitor is connected to another battery and 24.0 µC of charge is stored on the capacitor, what is the voltage of the battery? V(a) When a battery is connected to the plates of a 5.00-µF capacitor, it stores a charge of 60.0 µC. What is the voltage of the battery? V (b) If the same capacitor is connected to another battery and 75.0 µC of charge is stored on the capacitor, what is the voltage of the battery? V Need Help? Read It
- c. Charge on C₂ (in nC) = Submit Answer Incorrect. Tries 4/10 Previous Tries d. Voltage across C₂ (in V) = Submit Answer Tries 0/10 e. Charge on C3 (in nC) = Submit Answer Tries 0/10 f. Voltage across C3 (in V) =There are two parallel plate capacitors. They each have plates with areas of 2.3 meters squared, and plate separations of 9 mm. The first capacitor has a dielectric with a dielectric constant of 18 and the second has a dielectric with a dielectric constant of 19. What is the capacitance of these two capacitors in parallel with each other? Give your answer in microfarads.Please answer this question
- If a capacitor is connected to a battery, and allowed enough time to charge, the battery will maintain a constant voltage drop across the capacitor. This means that ∆V cannot change. It turns out that the effect on the capacitor due to inserting a dielectric is still the same; the capacitance increases by a factor of k. Since the voltage difference must stay the same, how can this happen? What must the battery do to create and maintain this voltage drop as the dielectric is inserted?A capacitor measures 5 cm by 8 cm, and its plates are separated by a 1 mm air gap. How much charge will this capacitor store when it has 12 V across it? O 1.27 x 10^-10 C O 4.25 x 10^-10 C O 9.3 x 10^-11 C O 3.2 x 10^-11 C 8.9 x 10^-9 C