Three capacitors each of capacitance 1.5\(\mu F\) are connected in series. The total capacitance in the circuit is
Answer Details
When capacitors are connected in series, the reciprocal of the total capacitance is equal to the sum of the reciprocals of the individual capacitances. Mathematically, we can express it as:
\(\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ...\)
In this case, we have three capacitors of equal capacitance, so we can simplify the equation to:
\(\frac{1}{C_{total}} = \frac{1}{1.5\mu F} + \frac{1}{1.5\mu F} + \frac{1}{1.5\mu F}\)
\(\frac{1}{C_{total}} = \frac{3}{1.5\mu F}\)
\(\frac{1}{C_{total}} = 2\)
\(C_{total} = \frac{1}{2}\mu F\)
Therefore, the total capacitance in the circuit is 0.5 \(\mu F\), which is option (D).