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Question
A device Y is connected across an AC source of emf e = e0 sin ωt. The current through Y is given as i = i0 sin (ωt + π/2).
- Identify the device Y and write the expression for its reactance.
- Draw graphs showing a variation of emf and current with time over one cycle of AC for Y.
- How does the reactance of the device Y vary with the frequency of the AC? Show graphically.
- Draw the phasor diagram for device Y.
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Solution
1. The device Y is a capacitor. Its reactance is `"X"_"C" = 1/(omega"C")`, where ω is the angular frequency of the applied emf and C is the capacitance of the capacitor.
2.
3. `"X"_"C" = 1/(omega"C") = 1/(2pi"fC")`. Thus `"X"_"C" prop 1/"f"`, where f is the frequency of AC.
Suppose C = `(1000/(2pi))`pF
For f = 100 Hz, XC = 1 × 107 Ω = 10 M Ω;
for f = 200 Hz, XC = 5 M Ω;
for f = 300 Hz, XC = `10/3` M Ω;
for f = 400 Hz, XC = 2.5 M Ω
for f = 500 Hz, XC = 2 M Ω and so on.
(1 M Ω = 106 Ω)
Variation of `1/(omega"C")` with f
4. Phasor diagram for a purely capacitive circuit
The phasor representing the peak emf (e0) makes an angle wt in an anticlockwise direction with respect to the horizontal axis. As the current leads the voltage by 90°, the phasor representing the peak current (i0) is turned 90° anticlockwise with respect to the phasor representing emf e0. The projections of these phasors on the vertical axis give instantaneous values of e and i.
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