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In the figure below, the capacitance of each capacitor is $3 \mu \mathrm{F}$. The effective capacitance between $A$ and $B$ is
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Verified Answer
The correct answer is:
$5 \mu \mathrm{F}$
From figure,
$C_1=C_2=C_3=C_4=3 \mu \mathrm{F} \quad (given)$
$\because C_4$ and $C_3$ are in parallel, i.e.
$C_{\text {eq }}=C_4+C_3=(3+3) \mu \mathrm{F}=6 \mu \mathrm{F}$
Now arrangement of capacitors will be as follows
$\because C_1$ is in parallel with the series combination of $C_2$ and $C_{\text {eq }}^{\prime}$.
$\therefore C_{\text {eq }}^{\prime \prime}$ between $A$ and $B$
$=\frac{\left(C_2 \times C_{\mathrm{eq}}^{\prime}\right)}{C_2+C_{\mathrm{eq}}^{\prime}}+C_1 \Rightarrow\left(\frac{3 \times 6}{3+6}\right)+3=5 \mu \mathrm{F}$
$C_1=C_2=C_3=C_4=3 \mu \mathrm{F} \quad (given)$
$\because C_4$ and $C_3$ are in parallel, i.e.
$C_{\text {eq }}=C_4+C_3=(3+3) \mu \mathrm{F}=6 \mu \mathrm{F}$
Now arrangement of capacitors will be as follows
$\because C_1$ is in parallel with the series combination of $C_2$ and $C_{\text {eq }}^{\prime}$.
$\therefore C_{\text {eq }}^{\prime \prime}$ between $A$ and $B$
$=\frac{\left(C_2 \times C_{\mathrm{eq}}^{\prime}\right)}{C_2+C_{\mathrm{eq}}^{\prime}}+C_1 \Rightarrow\left(\frac{3 \times 6}{3+6}\right)+3=5 \mu \mathrm{F}$
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