Question

Find effective capacity between a and b

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Answer 1

Key Points :

If capacitors are connected in series ,

$\bigstar\ \; \sf \pink{\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+...}$

If capacitors are connected in parallel ,

$\bigstar\ \; \sf \green{C_{eq}=C_1+C_2+C_3+...}$

Solution :

C₁ , C₂ are connected in series ,

$\sf \to \dfrac{1}{C_{12}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$

$\to \sf \dfrac{1}{C_{12}}=\dfrac{1}{2}+\dfrac{1}{2}$

$\to \sf \dfrac{1}{C_{12}}=\dfrac{2}{2}$

$\to \sf C_{12}=1 \mu F\ \; \bigstar$

C₁₂ , C₃ are connected in parallel ,

$\to \sf C_{123}=C_{12}+C_3$

$\to \sf C_{123}=1+1$

$\to \sf C_{123}=2\mu F\ \; \bigstar$

C₁₂₃ , C₄ are connected in series ,

$\to \sf \dfrac{1}{C_{1234}}=\dfrac{1}{C_{123}}+\dfrac{1}{C_4}$

$\to \sf \dfrac{1}{C_{1234}}=\dfrac{1}{2}+\dfrac{1}{2}$

$\to \sf \dfrac{1}{C_{1234}}=\dfrac{2}{2}$

$\to \sf \dfrac{1}{C_{1234}}=\dfrac{1}{1}$

$\to \sf C_{1234}=1 \mu F\ \; \bigstar$

C₁₂₃₄ , C₅ are connected in parallel ,

$\to \sf C_{12345}=C_{1234}+C_5$

$\sf \to C_{eq}=1+1$

$\leadsto \sf \blue{C_{AB}=2 \mu F}\ \; \bigstar$

Option (C)

Answer 2

Answer:-

$\longrightarrow \sf {C_{AB}=2 \mu F}\ \; \bigstar$

Option (C)

Correct explanation :-

C₁ , C₂ are connected in series ,

$\sf \to \dfrac{1}{C_{12}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$

$\to \sf \dfrac{1}{C_{12}}=\dfrac{1}{2}+\dfrac{1}{2}$

$\to \sf \dfrac{1}{C_{12}}=\dfrac{2}{2}$

$\to \sf C_{12}=1 \mu F\ \;$

C₁₂ , C₃ are connected in parallel ,

$\to \sf C_{123}=C_{12}+C_3$

$\to \sf C_{123}=1+1$

$\to \sf C_{123}=2\mu F\ \;$

C₁₂₃ , C₄ are connected in series ,

$\to \sf \dfrac{1}{C_{1234}}=\dfrac{1}{C_{123}}+\dfrac{1}{C_4}$

$\to \sf \dfrac{1}{C_{1234}}=\dfrac{1}{2}+\dfrac{1}{2}$

$\to \sf \dfrac{1}{C_{1234}}=\dfrac{2}{2}$

$\to \sf \dfrac{1}{C_{1234}}=\dfrac{1}{1}$

$\to \sf C_{1234}=1 \mu F\ \;$

C₁₂₃₄ , C₅ are connected in parallel ,

$\to \sf C_{12345}=C_{1234}+C_5$

$\sf \to C_{eq}=1+1$

$\longrightarrow \sf {C_{AB}=2 \mu F}\ \;$

Option (C)