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What is the effective capacitance of Three capacitance in which 2ñF and 6ñF connected in parallel then 2ñF is connected in series.
What is the total energy stored by the capacitors
wrong. The formulas for series/parallel capacitors are the opposite of that for resistors.
Parallel:
1/R = 1/R1 + 1/R2
C = C1 + C2
series:
R = R1 + R2
1/C = 1/C1 + 1/C2
You are correct. I apologize for the mistake. Let's calculate the effective capacitance again.
The capacitance of the 2µF and 6µF capacitors connected in parallel is simply the sum of the individual capacitances:
C_parallel = C1 + C2
C_parallel = 2µF + 6µF
C_parallel = 8µF
Next, we calculate the effective capacitance when this parallel combination is connected in series with the 2µF capacitor:
1/C_series = 1/C_parallel + 1/C3
1/C_series = 1/8µF + 1/2µF
1/C_series = 1/8µF + 4/8µF
1/C_series = 5/8µF
Taking the reciprocal of both sides, we find that the effective capacitance of the three capacitors is:
C_series = 8/5µF
Now, to calculate the total energy stored by the capacitors, we need to know the voltage applied across the capacitors. If the voltage is V, then the energy stored in each capacitor is given by:
E = (1/2) * C * V^2
The total energy stored by all the capacitors would be the sum of the energies of each capacitor:
Total energy = (1/2) * (2µF) * V^2 + (1/2) * (6µF) * V^2 + (1/2) * (8/5µF) * V^2
Simplifying, we get:
Total energy = V^2 [2/2 + 6/2 + 8/5]
Total energy = V^2 [1 + 3 + 8/5]
Total energy = V^2 [19/5]
Therefore, the total energy stored by the capacitors is (19/5) * V^2.
To find the effective capacitance of three capacitors, we need to calculate the equivalent capacitance of the capacitors connected in parallel and then combine it with the capacitance in series.
1. Capacitors in parallel:
When capacitors are connected in parallel, their capacitances add up. So, the equivalent capacitance (C_eq) of the 2ñF and 6ñF capacitors connected in parallel is:
C_eq = 2ñF + 6ñF
C_eq = 8ñF
2. Capacitors in series:
When capacitors are connected in series, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of each capacitance. In this case, the 2ñF capacitor is connected in series with the parallel combination of the 2ñF and 6ñF capacitors:
1 / C_total = 1 / (C_eq) + 1 / 2ñF
To simplify this equation, we first find the reciprocal of C_eq:
1 / C_eq = 1 / 8ñF
Substituting this value into the equation for the total capacitance:
1 / C_total = 1 / 8ñF + 1 / 2ñF
Finding the common denominator on the right side of the equation:
1 / C_total = (1 + 4) / (8ñF)
1 / C_total = 5 / (8ñF)
Taking the reciprocal of both sides to find C_total:
C_total = (8ñF) / 5
C_total = 1.6ñF
Therefore, the effective capacitance of the three capacitors is 1.6ñF.
Now, to find the total energy stored by the capacitors, we can use the formula:
Energy (E) = (1/2) * C_total * V^2
where V is the voltage across the capacitors (which we do not have in this question).
Please provide the voltage value so that we can calculate the total energy stored by the capacitors.
To find the effective capacitance, we first need to find the equivalent capacitance of the 2µF and 6µF capacitors connected in parallel. The formula for capacitors in parallel is:
1/C_parallel = 1/C1 + 1/C2
Plugging in the values, we get:
1/C_parallel = 1/2µF + 1/6µF
Simplifying, we get:
1/C_parallel = 3/6µF + 1/6µF
1/C_parallel = 4/6µF
1/C_parallel = 2/3µF
Taking the reciprocal of both sides, we find that the equivalent capacitance of the 2µF and 6µF capacitors in parallel is:
C_parallel = 3/2µF
Next, we need to find the effective capacitance when this parallel combination is connected in series with the 2µF capacitor. The formula for capacitors in series is simply the sum of the individual capacitances:
C_series = C_parallel + C2
C_series = 3/2µF + 2µF
C_series = 7/2µF
Therefore, the effective capacitance of the three capacitors is 7/2µF.
To find the total energy stored by the capacitors, we use the formula:
E = (1/2) * C * V^2
where E is the energy, C is the capacitance, and V is the voltage.
The voltage across the capacitors is not given in the question, so we cannot calculate the total energy without that information.