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

The Peak Voltage is 17 V

Explanation:

Given,  

E = 12 Volts DC

And We know that -  

• Heat Produced in a circuit when current passes through a circuit of R, resistance.

or H = $i^2RT$

So, Heat Produced Before and afterward -

$i^2RT = i_{rms}^{2}RT$

$\frac {E^2}{R^2} = \frac {E_{rms}^{2}}{R^2}$

$E^2 = \frac {E_{0}^{2}}{2}$

$E_{0}^{2} = 2E^2$

$E_{0}^{2} = 2\times 12^2 = 2 \times 144$

$E_0 = \sqrt 2\times \sqrt144$

16.97 $\approx$ 17 V