A 5.1 µF capacitor is charged by a 14.4 V battery and then is disconnected from the battery. When this capacitor is then connected to a second capacitor, the voltage on the first drops to 6 V.

Find the capacitance of the second capacitor.

The capacitance of the second capacitor is approximately 4.24 uF. Assuming the second capacitor is uncharged, we can use the principle of conservation of charge to determine its capacitance. The charge on the first capacitor before it is connected to the second capacitor is:

Q = C1 * V1 = (5.1 uF) * (14.4 V) = 73.44 uC

When the first capacitor is connected to the second capacitor, the charge is shared between the two capacitors. Let C2 be the capacitance of the second capacitor, and let V be the common voltage across the two capacitors. The total charge on the two capacitors is:

Q = C1 * V = (C1 + C2) * V

We can solve for C2:

C2 = Q/V - C1 = (73.44 uC)/(6 V) - 5.1 uF = 4.24 uF

Therefore, the capacitance of the second capacitor is approximately 4.24 uF.

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A 5.1 µF capacitor is charged by a 14.4 V battery and then is disconnected from the battery. When this capacitor is then connected to a second capacitor, the voltage on the first drops to 6 V. Find the capacitance of the second capacitor.

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A 5.1 µF capacitor is charged by a 14.4 V battery and then is disconnected from the battery. When this capacitor is then connected to a second capacitor, the voltage on the first drops to 6 V.

Find the capacitance of the second capacitor.