In a series connection solve for the total capacitance.
C1 = 3.5 x 10^-12 F
C2 = 2.41 x 10^-12 F
C3 = 1.13 x 10^-12 F
V = 1700 V
Solution;
Capacitance in series is given as;
1CT=1C1+1C2+1C3+....\frac{1}{C_T}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+....CT1=C11+C21+C31+....
By substitution;
1CT=13.5×10−12+12.41×10−12+11.13×10−12\frac{1}{C_T}=\frac{1}{3.5×10^{-12}}+\frac{1}{2.41×10^{-12}}+\frac{1}{1.13×10^{-12}}CT1=3.5×10−121+2.41×10−121+1.13×10−121
1CT=1.5856×1012\frac{1}{C_T}=1.5856×10^{12}CT1=1.5856×1012
CT=11.5856×10−12C_T=\frac{1}{1.5856×10^{-12}}CT=1.5856×10−121
CT=6.307×10−13C_T=6.307×10^{-13}CT=6.307×10−13 F
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