Question

Consider a $250W$ heat lamp fixed to the ceiling in a bathroom. If the filament in one light burns out then the remaining three still work. Construct a problem in which you determine the resistance of each filament in order to obtain a certain intensity projected on the bathroom floor. The ceiling is $3.0m$ high. The problem will need to involve concave mirrors behind the filaments. Your instructor may wish to guide you on the level of complexity to consider in the electrical components.

Short answer

The resistance of each filament to achieve $416.66W/m^2$ intensity the value is,$57.14\mathrm{\Omega }$ .

Step-by-step solution

Definition of Resistance of filament lamp

There is a large difference in resistance between the off and on states of an incandescent lamp when the filament heats up. A typical$\mathbf{60}\mathbf{W}$bulb working at$\mathbf{250}\mathbf{v}\mathbf{o}\mathbf{l}\mathbf{t}\mathbf{s}$will draw$\mathbf{0}.\mathbf{24}\mathbf{a}\mathbf{m}\mathbf{p}\mathbf{s}$and have a resistance of around$\mathbf{1041}$.

Given Data

• Power of the lamp$250W$.
• Height of the ceiling$3.0m$.

Formula used

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

$I={\displaystyle \frac{P}{V}}\dots \dots (1)$

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,

$R={\displaystyle \frac{V}{l}}\dots ...(2)$

The magnification is expressed as,

$m=-{\displaystyle \frac{d_i}{d_o}}\dots ...(3)$

Here,

-$d_i$is the image distance.

-$d_o$is the object distance.

The expression for the intensity of the radiation is,

$I={\displaystyle \frac{{\displaystyle \frac{1}{2}}P}{A_i}}\dots ...(4)$

Here $A_i$ is the area of the image.

Explanation of solution

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

Substitutevaluesin equation$(1)$,

$\begin{array}{r}I={\displaystyle \frac{250W}{120V}}\\ =2.1A\end{array}$

Substitutevaluesin equation$(2)$.

$\begin{array}{r}R={\displaystyle \frac{120V}{2.1A}}\\ =57.14\mathrm{\Omega }\end{array}$

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

$\begin{array}{r}m=-{\displaystyle \frac{3m}{(30cm)\left({\displaystyle \frac{1a}{{10}^{2}am}}\right)}}\\ =-10\end{array}$

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$300c{m}^2$for$A_o$in the above expression.

$\begin{array}{r}{A}_i=|-10|\left(300c{m}^2\times {\displaystyle \frac{1m^2}{{10}^{4}c{m}^2}}\right)\\ =0.300m^2\end{array}$

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

$\begin{array}{r}I={\displaystyle \frac{250W}{2\left(0300m^2\right)}}\\ =416.66W/m^2\end{array}$

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