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.
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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
3 answers
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
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.
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.