Given three capacitors 0.3uF, 0.5 uF and 0.2 uF, the joint capacitance when arranged to give minimum capacitance is?

in parallel, they just add up

in series, the resultant capacitance x is found by using
1/x = 1/.3 + 1/.5 + 1/.2
x = 0.097 uF

To find the joint capacitance when the capacitors are arranged to give minimum capacitance, we need to consider them in parallel configuration.

The formula to calculate the total capacitance for capacitors in parallel is:

1/C_total = 1/C1 + 1/C2 + 1/C3 + ...

Let's substitute the values of the capacitors given:

1/C_total = 1/0.3uF + 1/0.5uF + 1/0.2uF

To add fractions, we need to have a common denominator. In this case, we can use 1uF as the common denominator:

1/C_total = (1/0.3uF) * (1uF/1uF) + (1/0.5uF) * (1uF/1uF) + (1/0.2uF) * (1uF/1uF)

Simplifying the equation:

1/C_total = (1uF/0.3uF) + (1uF/0.5uF) + (1uF/0.2uF)
1/C_total = 3.333uF + 2uF + 5uF
1/C_total = 10.333uF

Taking the reciprocal of both sides to find C_total:

C_total = 1/10.333uF
C_total = 0.0967uF (approximately)

Therefore, when the capacitors are arranged to give minimum capacitance, the joint capacitance is approximately 0.0967uF.

To find the joint capacitance when the capacitors are arranged to give the minimum capacitance, first, we need to understand how capacitors are connected in parallel and in series.

1. Capacitors in Parallel: When capacitors are connected in parallel, the total capacitance, C_parallel, is the sum of the individual capacitances:

C_parallel = C1 + C2 + C3

2. Capacitors in Series: When capacitors are connected in series, the reciprocal of the total capacitance, 1/C_series, is the sum of the reciprocals of the individual capacitances:

1/C_series = 1/C1 + 1/C2 + 1/C3

Now, let's calculate the joint capacitance in both cases and compare them:

1. Parallel Arrangement:
C_parallel = 0.3uF + 0.5uF + 0.2uF
= 1uF (Microfarad)

2. Series Arrangement:
1/C_series = 1/0.3uF + 1/0.5uF + 1/0.2uF
= 10/3uF

Taking the reciprocal to find C_series:
C_series = 3/10uF

Comparing the two values, we see that the joint capacitance when arranged to give the minimum capacitance is C_series = 3/10uF (0.3uF).