Question
Take $\text{C}_1=4.0\mu\text{F}$ and $\text{C}_2=6.0\mu\text{F}$ in figure. Calculate the equivalent capacitance of the combination between the points indicated.
✓
Answer
- $\therefore C_1, C_1$ are series & $C_2, C_2 $ are series as the V is same at p & q. So no current pass through p & q.
$\frac{1}{\text{C}}=\frac{1}{\text{C}_1}=\frac{1}{\text{C}_2}$
$\Rightarrow\frac{1}{\text{C}}=\frac{1+1}{\text{C}_1\text{C}_2}$
$\text{C}_{\text{p}}=\frac{\text{C}_1}{2}=\frac{4}{2}=2\mu\text{F}$
And $\text{C}_{\text{q}}=\frac{\text{C}_2}{2}=\frac{6}{2}=3\mu\text{F}$
$\therefore\text{C}=\text{C}_{\text{p}}+\text{C}_{\text{q}}=2+3=5\mu\text{F}$
- $\text{C}_1=4\mu\text{F},\ \text{C}_2=6\mu\text{F}$
In case of p & q, q = 0
$\therefore\text{C}_{\text{p}}=\frac{\text{C}_1}{2}=\frac{4}{2}=2\mu\text{F}$
$\text{C}_{\text{q}}=\frac{\text{C}_2}{2}=\frac{6}{2}=3\mu\text{F}$
& $\text{C}'=2+3=5\mu\text{F}$
$\text{C}\ \&\ \text{C}' =5\mu\text{F}$
$\therefore$ The equation of capacitor $\text{C}=\text{C}'+\text{C}''=5+5=10\mu\text{F}$
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