QuestionJuly 26, 2025

1. Find the total capacitance of two capacitors a) in parallel (Cl)=10times 10^-9 and (C2)=.068times 10^-6 x 10-6 farads b) same capacitors in series.

1. Find the total capacitance of two capacitors a) in parallel (Cl)=10times 10^-9 and (C2)=.068times 10^-6 x 10-6 farads b) same capacitors in series.
1. Find the total capacitance of two capacitors a) in parallel (Cl)=10times 10^-9 and (C2)=.068times 10^-6 x 10-6 farads b) same capacitors in series.

Solution
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Answer

a) C_{\text{total, parallel}} = 7.8 \times 10^{-8} farads ### b) C_{\text{total, series}} = 9.86 \times 10^{-9} farads Explanation 1. Calculate total capacitance in parallel Capacitors in parallel add directly. Use C_{\text{total}} = C_1 + C_2. C_{\text{total}} = 10 \times 10^{-9} + 0.068 \times 10^{-6} farads. 2. Calculate total capacitance in series Capacitors in series use the formula \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}. \frac{1}{C_{\text{total}}} = \frac{1}{10 \times 10^{-9}} + \frac{1}{0.068 \times 10^{-6}}.

Explanation

1. Calculate total capacitance in parallel<br /> Capacitors in parallel add directly. Use $C_{\text{total}} = C_1 + C_2$. <br /> $C_{\text{total}} = 10 \times 10^{-9} + 0.068 \times 10^{-6}$ farads.<br /><br />2. Calculate total capacitance in series<br /> Capacitors in series use the formula $\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}$.<br /> $\frac{1}{C_{\text{total}}} = \frac{1}{10 \times 10^{-9}} + \frac{1}{0.068 \times 10^{-6}}$.
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