91Ó°ÊÓ

The resistance, or the ratio of voltage to current, for all or part of an electric circuit at a fixed temperature, is generally constant.

Ohm's law may be expressed mathematically as

$\frac{\mathrm{V}}{\mathrm{I}}=\mathrm{R}$………. (1)

Let the resistance of the bulb is and current flowing through the circuit is $I$.

And the potential of the battery is $\mathrm{Î\mu }$.

role="math" localid="1655785432702" $$\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \\ =\frac{\mathrm{Î\mu }}{\mathrm{R}+\mathrm{R}} \\ =\frac{\mathrm{Î\mu }}{2\mathrm{R}} \end{gathered}$$

Therefore, the potential difference across each bulb is

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}=\frac{\mathrm{Î\mu }}{2\mathrm{R}}\mathrm{R} \\ =\frac{\mathrm{Î\mu }}{2} \end{gathered}$$

And

$$\begin{gathered} {\mathrm{V}}_{\mathrm{B}}=\frac{\mathrm{Î\mu }}{2\mathrm{R}}\mathrm{R} \\ =\frac{\mathrm{Î\mu }}{2} \end{gathered}$$

So, the power consumption of each bulb is

$$\begin{gathered} {\mathrm{P}}_{\mathrm{A}}={\mathrm{V}}_{\mathrm{A}}\mathrm{I} \\ =\frac{\mathrm{Î\mu }}{2}·\frac{\mathrm{Î\mu }}{2\mathrm{R}} \\ {\mathrm{P}}_{\mathrm{A}}=\frac{{\mathrm{Î\mu }}^{\mathit{2}}}{4\mathrm{R}} \end{gathered}$$ ….. (2)

And

$$\begin{gathered} {\mathrm{P}}_{\mathrm{B}}={\mathrm{V}}_{\mathrm{B}}\mathrm{I} \\ =\frac{\mathrm{Î\mu }}{2}·\frac{\mathrm{Î\mu }}{2\mathrm{R}} \\ {\mathrm{P}}_{\mathrm{B}}=\frac{{\mathrm{Î\mu }}^{\mathit{2}}}{4\mathrm{R}} \end{gathered}$$ ….. (3)

Both equations (2) and (3) are same. Therefore,the bulb A glows just as brightly as the bulb B.

$$\begin{gathered} \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \\ =\frac{\mathrm{Î\mu }}{\mathrm{R}} \end{gathered}$$

And the potential difference will be

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}=\frac{\mathrm{Î\mu }}{\mathrm{R}}\mathrm{R} \\ =\mathrm{Î\mu } \end{gathered}$$

Therefore, now the power consumption will be

$\mathrm{P}=\mathrm{VI}$

$=\mathrm{Î\mu }·\frac{\mathrm{Î\mu }}{\mathrm{R}}$

$\mathrm{P}=\frac{{\mathrm{Î\mu }}^2}{\mathrm{R}}$ ….. (4)

Compare equations with previous one,you get

role="math" localid="1655786005802" $\mathrm{P}=4{\mathrm{P}}_{\mathrm{A}}$

Therefore, bulb A glows more brightly compared to its previous brightness.