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?

Answer 1

Voltage across the electric bulb, E = 12 volts
Let E₀ 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`

⇒ E₀² = 2 × 144

`⇒ E_0 = sqrt(2 xx 144)`

`⇒ E_0 = sqrt(2) xx 12`

`⇒ E_0 = 1.4142 xx 12`
⇒ E₀ = 16.9704
 = 16.97 ≈ 17 V

Thus , peak value of voltage is 17 V.