NEET-XII-Physics
39: Alternating Current
- #6The dielectric strength of air is 3.0 × 106 V/m. A parallel-plate air-capacitor has area 20 cm2 and plate separation 0.10 mm. Find the maximum rms voltage of an AC source that can be safely connected to this capacitor.Ans : Given:
Area of parallel-plate air-capacitor, A = 20 cm2
Separation between the plates, d = 0.1 mm
Dielectric strength of air, E= 3 × 106 V/m
E = `` \frac{V}{d}``,
where V = potential difference across the capacitor
`` \therefore `` V = Ed
= 3 × 106 × 0.1 × 10-3
= 3 × 102 = 300 V
Thus, peak value of voltage is 300 V.
Maximum rms value of voltage `` \left({V}_{\,\mathrm{\,rms\,}}\right)`` is given by,
`` {V}_{\,\mathrm{\,rms\,}}=\frac{{V}_{0}}{\sqrt{2}}``
`` =\frac{300}{\sqrt{2}}=212\,\mathrm{\,V\,}``
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