91Ó°ÊÓ

There is a large difference in resistance between the off and on states of an incandescent lamp when the filament heats up. A typical\({\bf{60W}}\)bulb working at\({\bf{250}}{\rm{ }}{\bf{volts}}\)will draw\({\bf{0}}.{\bf{24}}{\rm{ }}{\bf{amps}}\)and have a resistance of around\({\bf{1041}}\).

• Power of the lamp\(250\;W\).
• Height of the ceiling\(3.0\;m\).

The relation between the current and the power is given as,

\(I = \dfrac{P}{V} \ldots \ldots (1)\)

Here,

-\(P\)is the power of the lamp.

-\(V\)is the voltage applied to the lamp.

-\(I\)is the current.

The expression for the resistance is given as,

\(R = \dfrac{V}{l} \ldots ...{\rm{ }}(2)\)

The magnification is expressed as,

\(m = - \dfrac{{{d_i}}}{{{d_o}}} \ldots ...(3)\)

Here,

-\({d_i}\)is the image distance.

-\({d_o}\)is the object distance.

The expression for the intensity of the radiation is,

\(I = \dfrac{{\dfrac{1}{2}P}}{{{A_i}}} \ldots ...(4)\)

Here \({A_i}\) is the area of the image.

Consider the power of\(250\;\;W\)and a voltage of\(120\;V\)is applied across the parallel combination of four lamps when one lamp gets damaged. The distance between the object and the mirror is\(30\;\;cm\), and the distance between the mirror and the image is\(3.0\;\;m\). The area of the object\(300\;\;c{m^2}\).

Substitutevaluesin equation\((1)\),

\(\begin{aligned}I = \dfrac{{250\;W}}{{120\;\;V}}\\ = 2.1\;A\end{aligned}\)

Substitutevaluesin equation\((2)\).

\(\begin{aligned}R = \dfrac{{120\;V}}{{2.1\;\;A}}\\ = 57.14\;\Omega \end{aligned}\)

Substitutevaluesin equation\((3)\)to calculate the magnification of the object,

\(\begin{aligned}m = - \dfrac{{3\;\;m}}{{(30\;cm)\left( {\dfrac{{1a}}{{{{10}^2}am}}} \right)}}\\ = - 10\end{aligned}\)

The relation between the area of the object and the area of the image is given below.

\({A_i} = |m|{A_o}\)

Substitute\( - 10\)for\(m\)and\(300\;c{m^2}\)for\({A_o}\)in the above expression.

\(\begin{aligned}{A_i} = | - 10|\left( {300\;c{m^2} \times \dfrac{{1\;{m^2}}}{{{{10}^4}\;c{m^2}}}} \right)\\ = 0.300\;{m^2}\end{aligned}\)

Substitute\(250\;W\)for\(P\)and\(0.300\;{m^2}\)for\({A_i}\)in equation\((4)\)to calculate the intensity.

\(\begin{aligned}I = \dfrac{{250\;W}}{{2\left( {0300\;{m^2}} \right)}}\\ = 416.66\;W/{m^2}\end{aligned}\)

Therefore, the resistance and intensity values are \(57.14\;\Omega \) and \(416.66\;W/{m^2}\)respectively.