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NEET-XII-Physics

39: Alternating Current

with Solutions - page 5
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  • #6
    The 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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