Chapter 23, Problem 96A

(a)

To determine

The current through the bulb.

Answer to Problem 96A

The current through the bulb is $0.134\text{ A}$ .

Explanation of Solution

Given:

The voltage of battery is $V=1.50\text{ V}$ .

The internal resistance is $R_{\text{1}}=0.200\text{Ω}$ .

The resistance of bulb is $R_{\text{bulb}}=22.0\text{Ω}$ .

Formula used:

The expression for the resistance is,

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

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

Calculation:

The equivalent resistance is,

$\begin{array}{l}R=2R_1+R_{\text{bulb}}\\ R=2\times \left(0.200\Omega \right)+\left(22.0\Omega \right)\\ R=22.4\Omega \end{array}$

The current through the bulb is,

$\begin{array}{l}R=\frac{V}{I}\\ I=\frac{V}{R}\\ I=\frac{2\times 1.50\text{ V}}{22.4\Omega }\\ I=0.134\text{ A}\end{array}$

Conclusion:

Thus, the current through the bulb is $0.134\text{ A}$ .

(b)

To determine

The power used by the bulb.

Answer to Problem 96A

The power used by the bulb is $0.395\text{ W}$ .

Explanation of Solution

Given:

The voltage of battery is $V=1.50\text{ V}$ .

The internal resistance is $R_{\text{1}}=0.200\text{Ω}$ .

The resistance of bulb is $R_{\text{bulb}}=22.0\text{Ω}$ .

Formula used:

The expression for power is,

$P=I^{2}R$

Here, $I$ is the current and $R$ is the resistance.

Calculation:

Refer part (a).

The current through the bulb is $I=0.134\text{ A}$ .

The power used by the bulb is,

$\begin{array}{l}P=I^2{R}_{\text{bulb}}\\ P={\left(0.134\text{ A}\right)}^2\left(22.0\Omega \right)\\ P=0.395\text{ W}\end{array}$

Conclusion:

Thus, the power used by the bulb is $0.395\text{ W}$ .

(c)

To determine

The greater power if the batteries had no internal resistance.

Answer to Problem 96A

The greater power if the batteries had no internal resistance is $0.014\text{ W}$ .

Explanation of Solution

Given:

The voltage of battery is $V=1.50\text{ V}$ .

The internal resistance is $R_{\text{1}}=0.200\text{Ω}$ .

The resistance of bulb is $R_{\text{bulb}}=22.0\text{Ω}$ .

Formula used:

The expression for power is,

$P=\frac{V^2}{R}$

Here, $V$ is the voltage and $R$ is the resistance.

Calculation:

Refer part (b).

The power used by the bulb is $P_{\text{bulb}}=0.395\text{ W}$ .

The power used by the bulb if the batteries had no internal resistance is,

$\begin{array}{l}P=\frac{V^2}{R_{\text{bulb}}}\\ P=\frac{{\left(\text{2}\times \text{1}\text{.50 V}\right)}^2}{\left(22.0\Omega \right)}\\ P=0.409\text{ W}\end{array}$

The greater power used by the bulb if the batteries had no internal resistance is,

$\begin{array}{l}\Delta P=P-P_{\text{bulb}}\\ \Delta P=0.409\text{ W}-0.395\text{W}\\ \Delta P=0.014\text{ W}\end{array}$

Conclusion:

Thus, the greater power if the batteries had no internal resistance is $0.014\text{ W}$ .