Question

Obtain the equivalent capacitance of the network in Fig. 2.35.​

Answer 1

GiveN :

• C1 = 100 pF
• C2 = 200 pF
• C3 = 200 pF

To FinD :

• Equivalent capacitance

SolutioN :

For calculating the equivalent capacitance first we have to add capacitor C2 and C3 in series. And then Capacitor C1 in parallel with the equivalent capacitance of C2 and C3

First of All, add C2 and C3 in series :

$\large \: \: \bigstar \: \: \boxed{\sf{\dfrac{1}{C_s} \: = \: \dfrac{1}{C_2} \: + \: \dfrac{1}{C_3}}} \\ \\ \implies \sf{\dfrac{1}{C_{}} \: = \: \dfrac{1}{200} \: + \: \dfrac{1}{200}} \\ \\ \implies \sf{\dfrac{1}{C_{}} \: = \: \dfrac{2}{200}} \\ \\ \implies \sf{\dfrac{1}{C_{}} \: = \: \dfrac{1}{100}} \\ \\ \implies {\sf{C_{}\: = \: 100}} \\ \\ \underline{\sf{\therefore \: Capacitance \: of \: C_2 \: and \: C_3 \: is \: 100 \: pF}}$

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Now, add equivalent capacitance of C2 and C3 with C1 in parallel combination :

$\large \: \: \bigstar \: \: \boxed{\sf{C_{eq} \: = \: C_1 \: + \: C_{}}} \\ \\ \implies {\sf{C_{eq} \: = \: 100 \: + \: 100}} \\ \\ \implies \sf{C_{eq} \: = \: 200} \\ \\ \underline{\sf{\therefore \: Equivalent \: capacitance \: is \: 200 \: pF}}$