Question
Find the equivalence between A and B in the given circuit diagram.
✓
Answer
Sol. Due to the equivalence at A being in series
$\frac{1}{ C }=\frac{1}{ C _1}+\frac{1}{ C _2}$
$\frac{1}{C}=\frac{1}{2}+\frac{1}{2}=1 \quad \therefore C =1 \mu F$
Similarly at $B$, from $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$
$\frac{1}{ C }=\frac{1}{2}+\frac{1}{2}=1=1 \mu F$
$C =1 \mu F$
Therefore, capacitance between A and $B =1+1=2 \mu F$.
$\frac{1}{ C }=\frac{1}{ C _1}+\frac{1}{ C _2}$
$\frac{1}{C}=\frac{1}{2}+\frac{1}{2}=1 \quad \therefore C =1 \mu F$
Similarly at $B$, from $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$
$\frac{1}{ C }=\frac{1}{2}+\frac{1}{2}=1=1 \mu F$
$C =1 \mu F$
Therefore, capacitance between A and $B =1+1=2 \mu F$.
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