Question

A light bulb glows because it has resistance. The brightness of a light bulb increases with the electrical power dissipated in the bulb. (a) In the circuit shown in Fig. Q25.14a, the two bulbs A and B are identical. Compared to bulb A, does bulb B glow more brightly, just as brightly, or less brightly? Explain your reasoning. (b) Bulb B is removed from the circuit and the circuit is completed as shown in Fig. Q25.14b. Compared to the brightness of bulb A in Fig. Q25.14a, does bulb A now glow more brightly, just as brightly, or less brightly? Explain your reasoning

Short answer

(a) Bulb A glows just as brightly as bulb B.

(b) Now bulb A glows more brightly compared to its previous brightness.

Step-by-step solution

Ohm’s law

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)

Find the power consumption of each bulb

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

And the potential of the battery is $\epsilon$.

(a) According to Ohm’s law, you can define the current as below.

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

Therefore, the potential difference across each bulb is

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

And

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

So, the power consumption of each bulb is

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

And

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

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

Find the power consumption after removing one bulb

(b) Again by using Ohm’s law,you get

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

And the potential difference will be

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

Therefore, now the power consumption will be

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

$=\epsilon ·\frac{\epsilon }{\mathrm{R}}$

$\mathrm{P}=\frac{{\epsilon }^2}{\mathrm{R}}$ ….. (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.