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

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

Answer Details

When capacitors are arranged in parallel, the equivalent capacitance is given by the sum of individual capacitances. When capacitors are arranged in series, the inverse of the equivalent capacitance is given by the sum of the inverses of individual capacitances. To obtain the minimum capacitance, the capacitors should be arranged in series. Therefore, the equivalent capacitance will be: 1/C = 1/C1 + 1/C2 + 1/C3 Substituting the given values, we get: 1/C = 1/0.3μF + 1/0.5μF + 1/0.2μF Solving the above equation, we get: C = 0.086μF Therefore, the equivalent capacitance is approximately 0.1μF, which is the first option.