Question

a) What is the hot resistance of a $25\mathrm{W}$ light bulb that runs on $120\mathrm{V}$ AC? (b) If the bulb’s operating temperature is , what is its resistance at?

Short answer

1. Value of resistance$R_o$at time$T_o$is$R_0=576\mathrm{W}$.
2. Value of resistance at temperature is $3.2\times {10}^2\mathrm{W}$.

Step-by-step solution

Relation between power & potential difference and definition of resistance of conductor.

If a potential difference$\mathrm{\Delta }\mathrm{V}$is maintained across a circuit element, the power, or rate at which energy is supplied to the element, is

$\mathbf{P}\mathbf{=}\mathbf{I}\mathrm{\Delta }\mathbf{V}$

Because the potential difference across a resistor is given by$\mathrm{\Delta }\mathrm{V}=\mathrm{I}\mathrm{R}$,we can express the power delivered to a resistor as

$\mathbf{P}\mathbf{=}{\mathbf{I}}^{\mathbf{2}}\mathbf{R}$

$\mathbf{P}\mathbf{=}\frac{{\mathbf{(}\mathrm{\Delta }\mathbf{V}\mathbf{)}}^{\mathbf{2}}}{\mathbf{R}}$

The resistance$\mathbf{R}$of a conductor varies approximately linearly with temperature$\mathbf{T}$, as given below.

$\mathbf{R}\mathbf{=}{\mathbf{R}}_{\mathbf{0}}\left(\mathbf{1}\mathbf{+}\alpha \left(\mathbf{T}\mathbf{-}{\mathbf{T}}_{\mathbf{0}}\right)\right)$

Where${\mathrm{R}}_0$is the resistance at some reference temperature${\mathrm{T}}_0$and$\alpha$is the temperature coefficient of resistivity.

 Given data

The power consumption of the lightbulb at temperature$T_o$is:

$P_0=25\mathrm{W}$

The potential difference across the lightbulb is:

$\mathrm{\Delta }V=120\mathrm{V}$

The operating temperature of the lightbulb is:

$T_0=2700{}^{{}^{\circ }}\mathrm{C}$

The temperature coefficient of resistivity for tungsten is:$\alpha =4.5\times {10}^{-3}{}^{{}^{\circ }}{\mathrm{C}}^{-1}$.

Calculation of resistance ${\mathrm{R}}_0$of power-bulb using the power consumption at time ${\mathrm{T}}_0$.

(a)

The power consumption$P_o$of the light bulb at temperature$T_o$is found in terms of the potential difference$\mathrm{\Delta }V$and the resistance$R_o$at temperature$T_o$from Equation (1):

$P_0=\frac{{(\mathrm{\Delta }V)}^2}{R_0}$

Solve for${\mathrm{R}}_0$:

$R_o=\frac{{(\mathrm{\Delta }V)}^2}{P_o}$

Entering the values for$\mathrm{\Delta }\mathrm{V}$and${\mathrm{P}}_0$it gives:

$\begin{array}{c}{R}_0=\frac{{(120\mathrm{V})}^2}{25\mathrm{W}}\\ =576\mathrm{W}\end{array}$

Hence, the required result is $576\mathrm{W}$.

Calculation of resistance using the linear resistance equation.

(b)

The resistance of the lightbulb is found as a function of temperature change from Equation (2):

$R=R_0\left(1+\alpha \left(T-T_0\right)\right)$

For the given values of $R_0,\alpha ,T,$and$T_0$we obtain:

$\begin{array}{c}R=(576\mathrm{W})\left(1+\left(4.5\times {10}^{-3}{}^{{}^{\circ }}\mathrm{C}\right)\left(2600{}^{{}^{\circ }}\mathrm{C}-2700{}^{{}^{\circ }}\mathrm{C}\right)\right)\\ =3.2\times {10}^2\mathrm{W}\end{array}$

Therefore, resistance using the linear resistance equation is $3.2\times {10}^2\mathrm{W}$.