capacitor-8

Electricity and Magnetism Level 2

What is the equivalent capacitance of this circuit? \text{What is the equivalent capacitance of this circuit?}

1 4 p F \frac{1}{4} pF 3 4 p F \frac{3}{4} pF 3 2 p F \frac{3}{2} pF 2 3 p F \frac{2}{3} pF

Mohammad Khaza by Mohammad Khaza , 20, Bangladesh

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1 solution

Munem Shahriar
Feb 21, 2018

This is a series circuit, so

1 C eq = 1 C 1 + 1 C 2 + 1 C 3 \dfrac 1{C_{\text{eq}}} = \frac 1{C_1} + \frac 1{C_2} + \frac 1{C_3}

C eq = ( 1 C 1 + 1 C 2 + 1 C 3 ) − 1 . C_{\text{eq}} = \left( \frac 1{C_1} + \frac 1{C_2} + \frac 1{C_3}\right)^{-1}.

Where C 1 = C 2 = C 3 = 2 pF C_1 = C_2 = C_3 = 2 ~ \text{pF}

C eq = ( 1 2 + 1 2 + 1 2 ) − 1 = ( 3 2 ) − 1 = 2 3 pF . \begin{aligned} C_{\text{eq}} & = \left(\dfrac12 + \dfrac12 + \dfrac12 \right)^{-1} \\& = \left(\dfrac 3 2 \right)^{-1} \\& = \boxed{\dfrac 23} ~ \text{pF}.\\ \end{aligned}

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thanks for uploading a solution.

Mohammad Khaza - 3 years, 3 months ago

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