Calculate the total capacitance of a.5f,2f,6f arrange in parallel b.7f ,4f,3f arrange in series

1 answer

To calculate the total capacitance for capacitors arranged in parallel and series, we can use the following formulas:

a. Capacitors in Parallel

For capacitors in parallel, the total capacitance (\(C_{total}\)) is the sum of the individual capacitances:

\[ C_{total} = C_1 + C_2 + C_3 + \ldots \]

Given capacitors: 0.5 f, 2 f, and 6 f.

\[ C_{total} = 0.5 , \text{f} + 2 , \text{f} + 6 , \text{f} = 0.5 + 2 + 6 = 8.5 , \text{f} \]

b. Capacitors in Series

For capacitors in series, the total capacitance (\(C_{total}\)) is given by the reciprocal of the sum of the reciprocals of the individual capacitances:

\[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ldots \]

Given capacitors: 7 f, 4 f, and 3 f.

Calculating the reciprocal:

\[ \frac{1}{C_{total}} = \frac{1}{7 , \text{f}} + \frac{1}{4 , \text{f}} + \frac{1}{3 , \text{f}} \]

Finding a common denominator (which is 84):

\[ \frac{1}{C_{total}} = \frac{12}{84} + \frac{21}{84} + \frac{28}{84} = \frac{12 + 21 + 28}{84} = \frac{61}{84} \]

Now, take the reciprocal to find total capacitance:

\[ C_{total} = \frac{84}{61} \approx 1.377 , \text{f} \]

Summary:

  • a. Total capacitance in parallel: 8.5 f
  • b. Total capacitance in series: approximately 1.377 f