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 rms value is the root mean square value of the quantity. The rms value is the peak value of any quantity. The heat developed by the AC and by the DC source will be the same if the circuit has the same resistance and if the current is passed for the same interval of time.

Formula used: The formula of the heat produced is,
$\Rightarrow H = {i^2}RT$
Where the heat produced is H the current is I the resistance in the circuit is R and the time taken is T.
The formula of the relation between normal current and the root mean square value is equal to,
$\Rightarrow {E_{rms}} = \dfrac{{{E_o}}}{{\sqrt 2 }}$
The root mean square value of voltage is ${E_{rms}}$ and the normal voltage is${E_o}$.
The formula of the ohm’s law is given by,
$\Rightarrow E = I \cdot R$
Where e.m.f is V the current is I and the resistance is R.

Complete step by step solution:
It is given in the problem that an electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness we need to find the peak voltage of the source.
The heat produced by the DC source is equal to,
$\Rightarrow {H_1} = {i^2}RT$………eq. (1)
The heat generated by the AC source is equal to,
$\Rightarrow {H_2} = {i^2}_{rms}RT$………eq. (2)
The heat generated will be the same for both the sources if the time taken and the resistance will be the same, equating equation (1) and (2).
$\Rightarrow {H_1} = {H_2}$
$\Rightarrow {i^2}RT = {i^2}_{rms}RT$
$\Rightarrow {i^2} = {i^2}_{rms}$
$\Rightarrow i = {i_{rms}}$
By ohm’s law,
$\Rightarrow i = {i_{rms}}$
$\Rightarrow \dfrac{E}{R} = \dfrac{{{E_{rms}}}}{R}$
$\Rightarrow E = {E_{rms}}$………eq. (3)
The formula of the relation between normal current and the root mean square value is equal to,
$\Rightarrow {E_{rms}} = \dfrac{{{E_o}}}{{\sqrt 2 }}$………eq. (4)
The root mean square value of voltage is ${E_{rms}}$ and the normal voltage is ${E_o}$.
Replacing the value of equation (4) in equation (3).
$\Rightarrow E = {E_{rms}}$
$\Rightarrow E = \dfrac{{{E_o}}}{{\sqrt 2 }}$
$\Rightarrow {E_o} = \sqrt 2 \cdot E$
$\Rightarrow {E_o} = \sqrt 2 \times \left( {12} \right)$
$\Rightarrow {E_o} = 16 \cdot 97V$
$\Rightarrow {E_o} \approx 17V$

The peak voltage of the source is equal to ${E_o} = 17V$.

Note: It is advisable for the students to remember the formula of the ohm’s law, the formula of the heat generated and the relation between the normal and the peak value of any quantity. In the DC source the current is not changing but in the AC source the current is alternating.