Electrostatic Potential and Capacitance Test - 54

Given,

Both capacitors are in series

Capacitance of A & B, $${{C}_{A}}=2\mu F\,\,\And \,\,{{C}_{B}}=3\mu F$$

Total Potential is $${{V}_{T}}=({{V}_{A}}+{{V}_{B}})=10V$$

Let,

Potential difference on capacitor A & B, $${{V}_{A}},\,{{V}_{B}}$$  

In series, Capacitor have equal charge

$${{Q}_{A}}={{Q}_{B}}$$

$${{C}_{A}}{{V}_{A}}={{C}_{B}}{{V}_{B}}$$

$$\dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}}{{{C}_{A}}}$$

$$\dfrac{{{V}_{A}}+{{V}_{B}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}+{{C}_{A}}}{{{C}_{A}}}$$

$${{V}_{B}}=\dfrac{{{C}_{A}}({{V}_{A}}+{{V}_{B}})}{{{C}_{B}}+{{C}_{A}}}=\dfrac{2\times 10}{5}=4V$$

$${{V}_{B}}=4V$$

Similarly, $$\dfrac{{{V}_{A}}}{{{V}_{B}}}=\dfrac{{{C}_{B}}}{{{C}_{A}}}\Rightarrow \,\,\,{{V}_{A}}=\dfrac{{{V}_{B}}\times {{C}_{B}}}{{{C}_{A}}}\,\,$$

$$\Rightarrow \,{{V}_{A}}=\dfrac{4\times 3}{2}=6V$$

Potential difference on capacitor A & B, $${{V}_{A}}=6V,\,\,\,\,{{V}_{B}}=4V$$   

The given circuit can be made like above figure.

This forms a Wheatstone bridge so the capacitor of $5\,\mu F$gets removed. (fig.1)

Capacitors on the upper branch is in series so their equivalent is (fig.2)

$$ \dfrac{1}{{{C}_{eq}}}=\dfrac{1}{2}+\dfrac{1}{2} $$

$$ {{C}_{eq}}=1\,\mu F $$

Similarly, for the lower branch equivalent is $${{C}_{eq}}=1\,\mu F$$   (fig.2)

Capacitors  $$1\,\mu F$$ and $$1\,\mu F$$ are in parallel. So equivalent is $$2\,\mu F$$ (fig.3)

Now, $$2\,\mu F$$ and $$2\,\mu F$$ are in series.

So,

$$ \dfrac{1}{{{C}_{eq}}}=\dfrac{1}{2}+\dfrac{1}{2} $$

$$ {{C}_{eq}}=1\,\,\mu F $$