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Discharging of Capacitors

An electric b...

Question

An electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness , what would be the peak voltage of the source?

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Solution

We have,

Potential $= 12 v$

So, The rms value of the AC voltage must be equal to $12 v$ .

Thus, The peak value of the voltage is:

$= \sqrt{2} \times 12 v$

$= 16.85 v$

Thus, the peak voltage of the source is $= 16.85 v$

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