Question
A variable frequency a.c source is connected to a capacitor. How will the displacement current change with decrease in frequency?
✓
Answer
Capacitive reaction $\text{X}_\text{C}=\frac{1}{2\pi\text{fC}}$
Hence, $\text{X}_\text{C}\propto\frac{1}{\text{f}}$
As frequency decreases, XC increases and the conduction current is inversely proportional to $\text{X}_\text{C}\Big(\because\text{I}\propto\frac{1}{\text{X}_\text{C}}\Big)$.
It means the displacement current decreases as the conduction current is equal to the displacement current.
Hence, $\text{X}_\text{C}\propto\frac{1}{\text{f}}$
As frequency decreases, XC increases and the conduction current is inversely proportional to $\text{X}_\text{C}\Big(\because\text{I}\propto\frac{1}{\text{X}_\text{C}}\Big)$.
It means the displacement current decreases as the conduction current is equal to the displacement current.
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