Question

A series circuit consists of a 0.047 ÄμF capacitor, a 0.090 ÄμF capacitor, and a 420 V battery. Find the charge for the following situations.

(a) on each of the capacitors
______ Aμc (0.047 ÄμF capacitor)
______Aμc (0.090 ÄμF capacitor)
(b) on each of the capacitors if they are reconnected in parallel across the battery
______Äμc (0.047 ÄμF capacitor)
_______Äμc (0.090 ÄμF capacitor)

Answer 1

In a series circuit, the charge on each capacitor is the same. The charge on the 0.047 μF capacitor would be 19.74 μC and the charge on the 0.090 μF capacitor would be 37.80 μC. When the capacitors are reconnected in parallel, the charge on each capacitor will be different, but the total charge on each capacitor will remain the same.

Step-by-step explanation:

In a series circuit, the charge on each capacitor is the same. To calculate the charge on each capacitor, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. For the 0.047 μF capacitor, the charge would be Q = (0.047 μF)(420 V) = 19.74 μC. For the 0.090 μF capacitor, the charge would be Q = (0.090 μF)(420 V) = 37.80 μC.

If the capacitors are reconnected in parallel across the battery, the charge on each capacitor will be different. However, the total charge on each capacitor will remain the same. The charge on the 0.047 μF capacitor in parallel would be Q = (0.047 μF)(420 V) = 19.74 μC. The charge on the 0.090 μF capacitor in parallel would be Q = (0.090 μF)(420 V) = 37.80 μC.