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Alternating Current — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current5 Marks
Question
When a circuit element ‘X’ is connected across an a.c. source, a current of $\sqrt{2}\text{A}$ flows through it and this current is in phase with the applied voltage. When another element ‘Y’ is connected across the same a.c. source, the same current flows in the circuit but it leads the voltage by $\frac{\pi}{2}$ radians.
  1. Name the circuit element X and Y.
  2. Find the current that flows in the circuit when the series combination of X and Y is connected across the same a.c. voltage.
  3. Net impedance.
✓

Answer

  1. When circuit element is X, the current is in phase with the applied emf, this implies that X is pure resistance.
When circuit element Y is connected, the current leads the voltage by $\frac{\pi}{2}$ so Y is pure capacitance.
  1.  
Resistance of X = R; $\text{I}=\frac{\text{V}}{\text{R}}$

$\sqrt{2}=\frac{\text{V}}{\text{R}} \ ...(\text{i})$

Reactance of $\text{Y, X}_{\text{C}}=\frac{1}{\omega\text{C}},\text{I}=\frac{\text{V}}{\text{X}_{\text{C}}}\Rightarrow\sqrt{2\text{C}}=\frac{\text{V}}{\text{X}_{\text{C}}} \ ...(\text{ii})$

This implies $\text{X}_{\text{C}}=\text{R}$

When R and C are connected in series across same voltage source, then

Impedance $\text{Z}=\sqrt{\text{R}^2+\text{X}^2_{\text{C}}}=\sqrt{\text{R}^2+\text{R}^2}=\sqrt{2}\text{R} \ ...(\text{iii})$

$\therefore$ Current in circuit $\text{I}=\frac{\text{V}}{\text{Z}}=\frac{\text{V}}{\sqrt{2}\text{R}}$

From (1), $\frac{\text{V}}{\text{R}}=\sqrt{2},\therefore\text{I}=\frac{1}{\sqrt{2}}\times\sqrt{2}=1\text{A}$
  1.  
$\text{Z}=\sqrt{\text{R}^2+\Big(\frac{1}{\omega\text{C}}\Big)^2}$

i.e., impedance decreases with increase of angular frequency $\omega$

When $\omega=0,\text{Z}=\infty$

When $\omega=0,\text{Z = R}$

The graph of impedance Z versus $\omega$ is shows in fig.

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