Determine the total capacitance if a 30 pF capacitor is connected in series with 0.03µF capacitor.
Solution;
For capacitors in series;
1CT=1C1+1C2\frac{1}{C_T}=\frac{1}{C_1}+\frac{1}{C_2}CT1=C11+C21
1CT=130×10−12+10.03×10−6\frac{1}{C_T}=\frac{1}{30×10^{-12}}+\frac{1}{0.03×10^{-6}}CT1=30×10−121+0.03×10−61
CT=2.997×10−11FC_T=2.997×10^{-11}FCT=2.997×10−11F
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