Question
A device ‘X’ is connected to an ac source $\text{V = V}_0\sin\omega\text{t}.$ The variation of voltage, current and power in one cycle is show in the following graph:
  1. Identify the device ‘X’.
  2. Which of the curves, A, B and C represent the voltage, current and the power consumed in the circuit? Justify your answer.
  3. How does its impedance vary with frequency of the ac source? Show graphically.
  4. Obtain an expression for the current in the circuit and its phase relation with ac voltage.

Answer

  1. The device ‘X’ is a capacitor.
  2. Curve B: Voltage
Curve C: Current

Curve A: Power consumed in the circuit

Reason: This is because current leads the voltage in phase by $\frac{\pi}{2}$ for a capacitor.
  1. Impedance:
$\text{XC}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi}=\text{C}$

$\Rightarrow\text{X}_{\text{C}}\propto\frac{1}{\text{f}}$
  1.  


Voltage applied to the circuit is

$\text{V = V}_0\sin\omega\text{t}$

Due to this voltage, a charge will be produced which will charge the plates of the capacitor with positive and negative charges.

$\text{V}=\frac{\text{Q}}{\text{C}}\Rightarrow\text{Q}=\text{CV}$

Therefore, the instantaneous value of the current in the circuit is

$\text{I}=\frac{\text{dQ}}{\text{dt}}=\frac{\text{d(CV)}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{CV}_0\sin\omega\text{t})$

$\text{I}=\omega\text{CV}_0\cos\omega\text{t}=\frac{\text{V}_0}{\frac{1}{\omega\text{C}}}\sin\Big(\omega\text{t}+\frac{\pi}{2}\Big)$

$\text{I}=\text{I}_0\sin\Big(\omega\text{t}+\frac{\pi}{2}\Big)$

where, $\text{I}_0=\frac{\text{V}_0}{\frac{1}{\omega\text{C}}}$ = Peak value of current

Hence current leads the voltage in phase by $\frac{\pi}{2}.$

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