Chapter 27, Problem 1P

Two 1.50-V batteries—with their positive terminals in the same direction—are inserted in series into a flashlight. One battery has an internal resistance of 0.255 Ω, and the other has an internal resistance of 0.153 Ω. When the switch is closed, the bulb carries a current of 600 mA. (a) What is the bulb’s resistance? (b) What fraction of the chemical energy transformed appears as internal energy in the batteries?

To determine

The value of the bulb resistance.

Answer to Problem 1P

The value of the bulb resistance is $4.59\text{Ω}$.

Explanation of Solution

Given information: Internal resistance of one battery is $0.255\text{Ω}$, internal resistance of the other battery is $0.153\text{Ω}$, current when switch is closed is $600\text{mA}$, voltage across the batteries is $1.50\text{V}$.

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Write the expression for the total emf for the circuit.

$E=nV$ (1)

Here,

$E$ is the total emf for the circuit.

$n$ is the number of batteries.

$V$ is the voltage across the batteries.

Substitute $2$ for $n$, $1.50\text{V}$ for $V$ in equation (1) to find $E$,

$\begin{array}{c}E=2\times 1.50\text{V}\\ =3.0\text{V}\end{array}$

Thus, the total emf for the circuit is $3.0\text{V}$.

Formula to calculate the total internal resistance for the circuit.

$r=r_1+r_2$ (2)

Here,

$r$ is the total internal resistance for the circuit.

$r_1$ is the internal resistance of one battery

$r_2$ is the internal resistance of other battery

Substitute $0.255\text{Ω}$ for $r_1$, $0.153\text{Ω}$ for $r_2$ in equation (2) to find $r$,

$\begin{array}{c}r=0.255\text{Ω}+0.153\text{Ω}\\ =0.408\text{Ω}\end{array}$

Thus, the total internal resistance for the circuit is $0.408\text{Ω}$.

Formula to calculate the value of the bulb resistance.

$\begin{array}{c}E=I\left(R+r\right)\\ R=\frac{E}{I}-r\end{array}$ (3)

Here,

$I$ is the current when switch is closed.

$R$ is the resistance of the bulb.

Substitute $3.0\text{V}$ for $E$, $0.408\text{Ω}$ for $r$, $600\text{mA}$ for $I$ in equation (3) to find $R$,

$\begin{array}{c}R=\frac{3.0\text{V}}{600\text{mA}\times \frac{1\text{A}}{1000\text{mA}}}-0.408\text{Ω}\\ =4.59\text{Ω}\end{array}$

Thus, the value of the bulb resistance is $4.59\text{Ω}$.

Conclusion:

Therefore, the value of the bulb resistance is $4.59\text{Ω}$.

To determine

The fraction of the chemical energy transformed appears as internal energy in the batteries.

Answer to Problem 1P

The fraction of the chemical energy transformed appears as internal energy in the batteries is $8.16\%$.

Explanation of Solution

Given information: Internal resistance of one battery is $0.255\text{Ω}$, internal resistance of the other battery is $0.153\text{Ω}$, current when switch is closed is $600\text{mA}$, voltage across the batteries is $1.50\text{V}$.

Formula to calculate the internal energy of the battery.

$E^{\prime }=Ir$ (4)

Here,

$E^{\prime }$ is the internal energy of the battery.

Substitute $0.408\text{Ω}$ for $r$, $600\text{mA}$ for $I$ in equation (4) to find $E^{\prime }$,

$\begin{array}{c}{E}^{\prime }=\left(600\text{mA}\times \frac{1\text{A}}{1000\text{mA}}\right)\times 0.408\text{Ω}\\ =0.2448\text{V}\\ \approx \text{0}\text{.245}\text{V}\end{array}$

Thus, the internal energy of the battery is $\text{0}\text{.245}\text{V}$.

Formula to calculate the fraction of the chemical energy transformed appears as internal energy in the batteries.

$E_{\text{f}}=\frac{E^{\prime }}{E}\times 100$

Here,

$E_{\text{f}}$ is the fraction of the chemical energy transformed appears as internal energy in the batteries.

Substitute $3.0\text{V}$ for $E$, $\text{0}\text{.245}\text{V}$ for $E^{\prime }$ in equation (4) to find $E_{\text{f}}$,

$\begin{array}{c}{E}_{\text{f}}=\frac{\text{0}\text{.245}\text{V}}{3.0\text{V}}\times 100\\ =8.16\%\end{array}$

Thus the fraction of the chemical energy transformed appears as internal energy in the batteries is $8.16\%$.

Conclusion:

Therefore, the fraction of the chemical energy transformed appears as internal energy in the batteries is $8.16\%$.