Sum

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

Voltage across the electric bulb, *E* = 12 volts

Let *E*_{0} be the peak value of voltage.

We know that heat produced by passing an alternating current ( i ) through a resistor is equal to heat produced by passing a constant current `(i_{rms})`through the same resistor. If *R* is the resistance of the electric bulb and *T* is the temperature, then

`i^2RT = i^2_rms^{RT}`

`⇒ E^2/R^2 = E_{rms}^2/R^2`

`⇒ E^2 = E_0^2/2 (therefore E^2_rms = E_0^2)`

`⇒ E_0^2 = 2E^2`

`⇒ E_0^2 = 2xx(12)^2 = 2 xx 144`

`⇒ E_0= sqrt2xx144`

=16.97 = 17 ⇒ V

Thus , peak value of voltage is 17 V.

Concept: Peak and Rms Value of Alternating Current Or Voltage

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