The total capacitance is 44 F
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Given:
[tex]C_1\:=\:5.0\:F[/tex]
[tex]C_2\:=\:20.0\:F[/tex]
[tex]C_3\:=\:40.0\:F[/tex]
[tex]\\[/tex]
Required:
Total Capacitance [tex]C_T[/tex]
[tex]\\[/tex]
Equation:
Total Capacitance in Series Connection
[tex]\frac{1}{C_T}\:=\:\frac{1}{C_1}\:+\:\frac{1}{C_2}\:+\:\frac{1}{C_3}\:+\:...\:+\:\frac{1}{C_n}[/tex]
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Total Capacitance in Parallel Connection
[tex]C_T\:=\:C_1\:+\:C_2\:+\:C_3\:+\:...\:+\:C_n[/tex]
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Solution:
Note: Please refer to the photo for the illustration.
Solve for the Total Capacitance of [tex]C_1[/tex] and [tex]C_2[/tex] connected in Series connection
[tex]\frac{1}{C_{T12}}\:=\:\frac{1}{C_1}\:+\:\frac{1}{C_2}\:+\:\frac{1}{C_3}\:+\:...\:+\:\frac{1}{C_n}[/tex]
[tex]\frac{1}{C_{T12}}\:=\:\frac{1}{20\:F}\:+\:\frac{1}{5\:F}[/tex]
[tex]\frac{1}{C_{T12}}\:=\:\frac{1\:+\:4}{20\:F}[/tex]
[tex]\frac{1}{C_{T12}}\:=\:\frac{5}{20\:F}[/tex]
[tex]C_{T12}\:=\:\frac{20\:F}{5}[/tex]
[tex]C_{T12}\:=\:4\:F[/tex]
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Solve for the Total Capacitance of [tex]C_{T12}[/tex] and [tex]C_3[/tex] connected in Parallel connection
[tex]C_T\:=\:C_{T12}\:+\:C_3[/tex]
[tex]C_T\:=\:4.0\:F\:+\:40.0\:F[/tex]
[tex]C_T\:=\:44\:F[/tex]
[tex]\\[/tex]
Final Answer:
The total capacitance is 44 F
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