Question
Calculate the total capacitance of 5f,2f and 6f arrange in parallel
Answers
GPT-4o mini
When capacitors are arranged in parallel, the total capacitance is simply the sum of the individual capacitances.
Given the capacitances:
- \( C_1 = 5 \, \text{fF} \)
- \( C_2 = 2 \, \text{fF} \)
- \( C_3 = 6 \, \text{fF} \)
The total capacitance \( C_{\text{total}} \) can be calculated as follows:
\[
C_{\text{total}} = C_1 + C_2 + C_3
\]
Substituting the values:
\[
C_{\text{total}} = 5 \, \text{fF} + 2 \, \text{fF} + 6 \, \text{fF} = 13 \, \text{fF}
\]
Thus, the total capacitance of the capacitors in parallel is \( 13 \, \text{fF} \).
Given the capacitances:
- \( C_1 = 5 \, \text{fF} \)
- \( C_2 = 2 \, \text{fF} \)
- \( C_3 = 6 \, \text{fF} \)
The total capacitance \( C_{\text{total}} \) can be calculated as follows:
\[
C_{\text{total}} = C_1 + C_2 + C_3
\]
Substituting the values:
\[
C_{\text{total}} = 5 \, \text{fF} + 2 \, \text{fF} + 6 \, \text{fF} = 13 \, \text{fF}
\]
Thus, the total capacitance of the capacitors in parallel is \( 13 \, \text{fF} \).