Chapter 23, Problem 47A

To determine

The current through the bulb using Kirchhoff’s loop rule.

Answer to Problem 47A

The current through the bulb is $0.10\text{ A}$ .

Explanation of Solution

Given:

The resistance of the bulb is $R=15\Omega$ .

Formula used:

The expression for the resistance is,

$R=\frac{V}{I}$

Here, $V$ is the voltage and $I$ is current.

Calculation:

Given that a flashlight has three batteries, in which one of the batteries is put in backward.

The potential increase in battery, using Kirchoff’s loop rule, is

$\begin{array}{l}V=1.5\text{ V}+1.5\text{ V}-1.5\text{ V}\\ V=1.5\text{ V}\end{array}$

The current through the bulb is,

$\begin{array}{l}I=\frac{V}{R}\\ I=\frac{\text{1}\text{.5 V}}{\text{15 }\Omega }\\ I=0.10\text{ A}\end{array}$

Conclusion:

Thus, the current through the bulb is $0.10\text{ A}$ .