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AC Circuits — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsAC Circuits4 Marks
Question
A device $Y$ is connected across an $AC$ source of emf $e = e_0$ sinωt. The current through $Y$ is given as $i = i_0 \sin(\omega t + \pi /2)$
a) Identify the device $Y$ and write the expression for its reactance.
b) Draw graphs showing variation of emf and current with time over one cycle of AC for $Y$.
c) How does the reactance of the device $Y$ vary with the frequency of the $AC$ ? Show graphically
d) Draw the phasor diagram for the device $Y$.
✓

Answer

(a) The device $Y$ is a capacitor. Its reactance is $X_c=\frac{1}{\omega C}$, where $\omega$ is the angular frequency of the applied emf and $C$ is the capacitance of the capacitor.
(b)

Image
(c) $X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}$. Thus $X_C \propto \frac{1}{f}$, where $f$ is the frequency of $A C$. Suppose $C=\left(\frac{1000}{2 \pi}\right)$
$pF$
For $f =100 Hz , X _C=1 \times 10^7 \Omega=10 M \Omega$;
for $f=200 Hz , X_C=5 M \Omega$;
for $f=300 Hz , X _{ C }=\frac{10}{3} M \Omega$;
for $f=400 Hz , X _{ C }=2.5 M \Omega$
for $f=500 Hz , X_C=2 M \Omega$ and so on
$\left(1 M \Omega=10^6 \Omega\right)$.


Image
(d)


Image
The phasor representing the peak emf $\left(e_0\right)$ makes an angle $(\omega t)$ in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by $90^{\circ}$, the phasor representing the peak current ( $i _0$ ) is turned $90^{\circ}$ anticlockwise with respect to the phasor representing emf $e _0$. The projections of these phasors on the vertical axis give instantaneous values of $e$ and i .

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