Question

For an ideal battery$(\mathrm{r}=0)$, closing the switch in FIGURE P28.54 does not affect the brightness of bulb A. In practice, bulb A dims just a little when the switch closes. To see why, assume that the$1.50\mathrm{V}$battery has an internal resistance$\mathrm{r}=0.50\Omega$and that the resistance of a glowing bulb is$\mathrm{R}=6.00\Omega$.

a. What is the current through bulb A when the switch is open?

b. What is the current through bulb A after the switch has closed?

c. By what percentage does the current through A change when the switch is closed?

Short answer

a. when switch is open current though A is $0.23\mathrm{A}.$

b. when switch is closed current through A is $0.21\mathrm{A}.$

c. percentage change in the current is$8.7\%.$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

$$\begin{gathered} \mathrm{Emf}=1.5\mathrm{V} \\ \mathrm{Internal}\mathrm{resistance}=0.5\Omega \\ \mathrm{Resistance}\mathrm{of}\mathrm{bulb}=6\Omega \end{gathered}$$

We have to the current through bulb A when the switch is open.

Simplify

Using Kirchhoff voltage in the loop, Let current i is flowing.

$$\begin{gathered} \sum _{\mathrm{i}}V=0 \\ \mathrm{E}=\mathrm{i}(\mathrm{R}+\mathrm{r}) \\ \mathrm{i}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}=\frac{1.5\mathrm{V}}{6\Omega +0.5\Omega } \\ \mathrm{i}={\mathrm{I}}_1=0.23\mathrm{A} \end{gathered}$$

Part (b) Step 1: Given information

we have to find the current through bulb A after the switch has closed.

Simplify

The total resistance due to bulb will be

$$\begin{gathered} \frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{R_A}+\frac{1}{R_B} \\ \frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{6\Omega }+\frac{1}{6\Omega }=\frac{2}{6} \\ {R}_{eq}=3\Omega \end{gathered}$$

then current through the battery will be

$$\begin{gathered} {\mathrm{I}}_{\mathrm{BATTERY}}=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{eq}}+\mathrm{r}} \\ =\frac{1.5\mathrm{V}}{3\Omega +0.5\Omega } \\ =\frac{1.5\mathrm{V}}{3.5\Omega } \\ {\mathrm{i}}_{\mathrm{BATTERY}}=0.43\mathrm{A} \end{gathered}$$

Now using the Kirchhoff voltage rule in left loop

$$\begin{gathered} \underset{\mathrm{i}}{\sum \mathrm{V}=0} \\ -1.5\mathrm{V}+0.43\mathrm{A}\times 0.5\Omega +{\mathrm{I}}_{\mathrm{A}}\times 6\Omega =0 \\ {\mathrm{I}}_{\mathrm{A}}=\frac{1.5\mathrm{V}-2.15\mathrm{V}}{6\Omega } \\ {\mathrm{I}}_{\mathrm{A}}={\mathrm{I}}_2=0.214\mathrm{A} \end{gathered}$$

Part (c) Step 1: Given information

we have to find by what percentage does the current through A change when the switch is closed.

Simplify

The change in the current when switch is closed

$$\begin{gathered} ∆\mathrm{I}=\left(\frac{{\mathrm{I}}_1-{\mathrm{I}}_2}{{\mathrm{I}}_1}\right)\times 100\% \\ ∆\mathrm{I}=\frac{0.23-0.21}{0.23}\times 100\%=8.7\% \end{gathered}$$