A 300v battery is connected across capacitors 3uF and 6uF in parallel. Evaluate the charge and energy stored in each capacitor

I want numerical solution

We are given C1 and C2. V is the same across each capacitor, 300 V since they are in parallel.

Q = CV for each capacitor.
W = KE = 1/2 CV^2 for each capacitor

To find the charge and energy stored in each capacitor, we can use the formulas:

Q = CV
E = (1/2)CV^2

where Q is the charge, C is the capacitance, V is the voltage, and E is the energy stored.

Given:
Battery voltage (V) = 300 V
Capacitor 1 capacitance (C1) = 3 µF
Capacitor 2 capacitance (C2) = 6 µF

First, let's calculate the charge stored in each capacitor:

For capacitor 1:
Q1 = C1 * V
Q1 = 3 µF * 300 V
Q1 = 900 µC

For capacitor 2:
Q2 = C2 * V
Q2 = 6 µF * 300 V
Q2 = 1800 µC

Next, let's calculate the energy stored in each capacitor:

For capacitor 1:
E1 = (1/2) * C1 * V^2
E1 = (1/2) * 3 µF * (300 V)^2
E1 = (1/2) * 3 µF * 90000 V^2
E1 = 135000 µJ

For capacitor 2:
E2 = (1/2) * C2 * V^2
E2 = (1/2) * 6 µF * (300 V)^2
E2 = (1/2) * 6 µF * 90000 V^2
E2 = 270000 µJ

Therefore, the charge stored in capacitor 1 is 900 µC and in capacitor 2 is 1800 µC. The energy stored in capacitor 1 is 135000 µJ and in capacitor 2 is 270000 µJ.

To evaluate the charge and energy stored in each capacitor, we can follow these steps:

Step 1: Determine the total capacitance when capacitors are connected in parallel.
When capacitors are connected in parallel, their total capacitance (Ct) is given by the sum of individual capacitances. In this case, Ct = C1 + C2, where C1 is the capacitance of the 3uF capacitor and C2 is the capacitance of the 6uF capacitor.

Ct = 3uF + 6uF
Ct = 9uF

So, the total capacitance when the capacitors are connected in parallel is 9uF.

Step 2: Calculate the charge on each capacitor.
To find 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 across the capacitor.

For the 3uF capacitor:
Q1 = C1 * V
Q1 = 3uF * 300V
Q1 = 900uC (microcoulombs)

For the 6uF capacitor:
Q2 = C2 * V
Q2 = 6uF * 300V
Q2 = 1800uC (microcoulombs)

Therefore, the charge on the 3uF capacitor is 900 microcoulombs, and the charge on the 6uF capacitor is 1800 microcoulombs.

Step 3: Calculate the energy stored in each capacitor.
The energy stored in a capacitor (E) is given by the formula E = 1/2 * C * V^2, where E is the energy, C is the capacitance, and V is the voltage across the capacitor.

For the 3uF capacitor:
E1 = 1/2 * C1 * V^2
E1 = 1/2 * 3uF * (300V)^2
E1 = 135,000uJ (microjoules)

For the 6uF capacitor:
E2 = 1/2 * C2 * V^2
E2 = 1/2 * 6uF * (300V)^2
E2 = 540,000uJ (microjoules)

Therefore, the energy stored in the 3uF capacitor is 135,000 microjoules, and the energy stored in the 6uF capacitor is 540,000 microjoules.