What is the effective capacitance of Three capacitance in which 2ñF and 6ñF connected in parallel then 2ñF is connected in series.
In parallel connection, the effective capacitance (Cp) is equal to the sum of the individual capacitances:
Cp = 2µF + 6µF = 8µF
In series connection, the effective capacitance (Cs) is given by the reciprocal of the sum of the reciprocals of the individual capacitances:
Cs = 1 / (1/2µF + 1/8µF) = 1 / (4/8µF + 1/8µF) = 1 / (5/8µF) = 8µF / 5 = 1.6µF
Therefore, the effective capacitance of the three capacitances is 1.6µF.
To find the effective capacitance, first we need to calculate the equivalent capacitance of the two capacitors connected in parallel and then connect it in series with the third capacitor.
1. Capacitors Connected in Parallel (2uF and 6uF):
The formula for calculating the equivalent capacitance for capacitors connected in parallel is:
1/C_eq = 1/C_1 + 1/C_2 + ...
Let's substitute the given values:
1/C_eq = 1/2uF + 1/6uF
To add the fractions, we need to find a common denominator:
1/C_eq = (3/6uF + 1/6uF) / (3/6uF)
Simplifying the numerator:
1/C_eq = (4/6uF) / (3/6uF)
Now dividing the fractions:
1/C_eq = 4/3uF
Taking the reciprocal, we get:
C_eq = 3/4uF = 0.75uF
The equivalent capacitance of the two capacitors connected in parallel is 0.75uF.
2. Capacitor Connected in Series (0.75uF and 2uF):
The formula for calculating the equivalent capacitance for capacitors connected in series is:
C_eq = C_1 + C_2 + ...
Substituting the given values:
C_eq = 0.75uF + 2uF
Adding the capacitances:
C_eq = 2.75uF
The effective capacitance of the three capacitors, 2uF and 6uF connected in parallel and then 2uF connected in series, is 2.75uF.