Chapter 27, Problem 45AP

An ideal voltmeter connected across a certain fresh 9-V battery reads 9.30 V, and an ideal ammeter briefly connected across the same battery reads 3.70 A. We say the battery has an open-circuit voltage of 9.30 V and a short-circuit current of 3.70 A. Model the battery as a source of emf ε in series with an internal resistance r as in Figure 27.1a. Determine both (a) ε and (b) r. An experimenter connects two of these identical batteries together as shown in Figure P27.45. Find (c) the open-circuit voltage and (d) the short-circuit current of the pair of connected batteries. (e) The experimenter connects a 12.0-Ω resistor between the exposed terminals of the connected batteries. Find the current in the resistor. (f) Find the power delivered to the resistor. (g) The experimenter connects a second identical resistor in parallel with the first. Find the power delivered to each resistor. (h) Because the same pair of batteries is connected across both resistors as was connected across the single resistor, why is the power in part (g) not the same as that in part (f)?

Figure P27.45

To determine

The emf of the battery.

Answer to Problem 45AP

The emf of the battery is $9.30\text{V}$.

Explanation of Solution

Given info: The open circuit voltage of the battery is $9.30\text{V}$. The short circuit current is $3.70\text{A}$.

In an open circuit current the current of the battery is $0\text{A}$.

Formula to calculate the emf of the battery is,

$\epsilon =V+IR$

Here,

$\epsilon$ is the emf of the battery.

$V$ is the voltage in the battery.

$I$ is the current produced in the battery.

$R$ is the resistance in the battery.

Substitute $9.30\text{V}$  for $V$, $0\text{A}$ for $I$ in the above equation.

$\begin{array}{c}\epsilon =9.30\text{V}+\left(0\text{A}\right)\left(R\right)\\ =9.30\text{V}\end{array}$

Conclusion:

Therefore, the emf of the battery is $9.30\text{V}$.

To determine

The resistance of the battery.

Answer to Problem 45AP

The resistance of the battery is $2.51\Omega$.

Explanation of Solution

Given info: The open circuit voltage of the battery is $9.30\text{V}$. The short circuit current is $3.70\text{A}$.

Formula to calculate the internal resistance of the battery is,

$\Delta V=V-Ir=0$

Here,

$\Delta V$ is the voltage difference of the battery.

$I$ is the current produced in the battery.

$r$ is the internal resistance in the battery.

Substitute $9.30\text{V}$  for $V$, $3.70\text{A}$ for $I$ in the above equation.

$\begin{array}{c}9.30\text{V}-\left(3.70\text{A}\right)r=0\\ 9.30\text{V}=\left(3.70\text{A}\right)r\\ r=\frac{9.30\text{V}}{3.70\text{A}}\\ =2.51\Omega \end{array}$

Conclusion:

Therefore, resistance of the battery is $2.51\Omega$.

To determine

The open circuit voltage of the battery.

Answer to Problem 45AP

The open circuit voltage of the battery is $18.6\text{V}$.

Explanation of Solution

Given info: The open circuit voltage of the battery is $9.30\text{V}$. The short circuit current is $3.70\text{A}$. The batteries are connected in series. The voltages of the both batteries are same.

Formula to calculate the total emf of the battery is,

${\epsilon }_{\text{T}}=2\epsilon$

Here,

${\epsilon }_{\text{T}}$ is the total emf of the battery.

Substitute $9.30\text{V}$  for $\epsilon$ in the above equation.

$\begin{array}{c}{\epsilon }_{\text{T}}=2\left(9.30\text{V}\right)\\ =18.6\text{V}\end{array}$

The total emf of the battery is equal to the open circuit voltage of the battery.

$V_{\text{T}}=18.6\text{V}$

Conclusion:

Therefore, the open circuit voltage of the battery is $18.6\text{V}$.

To determine

The short circuit current of the pair of connected batteries.

Answer to Problem 45AP

The short circuit current of the pair of connected batteries is $3.70\text{A}$.

Explanation of Solution

Given info: The open circuit voltage of the battery is $9.30\text{V}$. The short circuit current is $3.70\text{A}$. The batteries are connected in series. The voltages of the both batteries are same.

The total resistance in the battery is,

$r_{\text{T}}=2r$

Here,

$r_{\text{T}}$ is the total resistance in the battery.

Substitute $2.51\Omega$ for $r$ in the above equation.

$\begin{array}{c}{r}_{\text{T}}=2\left(2.51\Omega \right)\\ =5.02\Omega \end{array}$

Thus, the internal resistance of the battery is $5.02\Omega$.

Formula to calculate the short circuit current of the batteries is,

$I=\frac{V_{\text{T}}}{r_{\text{T}}}$

Here,

$V_{\text{T}}$ is the open circuit voltage of the battery.

Substitute $18.6\text{V}$  for $V_{\text{T}}$, $5.02\Omega$ for $r_{\text{T}}$ in the above equation.

$\begin{array}{c}I=\frac{18.6\text{V}}{5.02\Omega }\\ =3.70\text{A}\end{array}$

Conclusion:

Therefore, the short circuit current of the pair of connected batteries is $3.70\text{A}$.

To determine

The current in the resistor if the $12.0\Omega$ resistor is connected between the exposed terminals of the connected battery.

Answer to Problem 45AP

The current in the resistor $12.0\Omega$ is $1.09\text{A}$.

Explanation of Solution

Given info: The open circuit voltage of the battery is $9.30\text{V}$. The short circuit current is $3.70\text{A}$. The batteries are connected in series. The voltages of the both batteries are same.

The total series resistance in the battery is,

$R_{\text{eq}}=r_{\text{T}}+12.0\Omega$

Here,

$R_{\text{eq}}$ is the total series resistance in the battery.

Substitute $5.02\Omega$ for $r_{\text{T}}$ in the above equation.

$\begin{array}{c}{R}_{\text{eq}}=5.02\Omega +12.0\Omega \\ =17.02\Omega \end{array}$

Thus, the total series resistance of the battery is $17.02\Omega$.

Formula to calculate the current in the resistor $12.0\Omega$ is,

$I_{\text{1}}=\frac{V_{\text{T}}}{R_{\text{eq}}}$

Here,

$I_1$ is the current in the resistor $12.0\Omega$.

Substitute $18.6\text{V}$  for $V_{\text{T}}$, $17.02\Omega$ for $R_{\text{eq}}$ in the above equation.

$\begin{array}{c}{I}_{\text{1}}=\frac{18.6\text{V}}{17.02\Omega }\\ =1.092\text{A}\\ \approx \text{1}\text{.09}\text{A}\end{array}$

Conclusion:

Therefore, the current in the resistor $12.0\Omega$ is $1.09\text{A}$.

To determine

The power delivered to the resistor.

Answer to Problem 45AP

The power delivered to the resistor is $14.3\text{W}$.

Explanation of Solution

Given info: The open circuit voltage of the battery is $9.30\text{V}$. The short circuit current is $3.70\text{A}$. The batteries are connected in series. The voltages of the both batteries are same.

Formula to calculate the power delivered to the resistor is,

$P={\left(I_1\right)}^{2}R$

Here,

$P$ is the power delivered to the resistor.

$R$ is the resistance of the resistor.

Substitute $1.09\text{A}$ for $I_1$, $12.0\Omega$ for $R$ in the above equation.

$\begin{array}{c}P={\left(1.09\text{A}\right)}^{2}12.0\Omega \\ =14.25\text{W}\\ \approx \text{14}\text{.3}\text{W}\end{array}$

Conclusion:

Therefore, the power delivered to the resistor is $14.3\text{W}$.

To determine

The power delivered to each resistor.

Answer to Problem 45AP

The power delivered to each resistor is $8.54\text{W}$.

Explanation of Solution

Given info: The open circuit voltage of the battery is $9.30\text{V}$. The short circuit current is $3.70\text{A}$.

The batteries are connected in series. The voltages of the both batteries are same.

The equivalent internal resistance in the battery is,

$\begin{array}{c}{R}_1=R\text{'}+r_{\text{T}}\\ =\left[\left(\frac{1}{R}\right)+\left(\frac{1}{R}\right)\right]+r_{\text{T}}\end{array}$

Here,

$R_1$ is the total resistance of the resistor.

$R\text{'}$ is the equivalent internal resistance in parallel.

Substitute $12.0\Omega$ for $R$, $5.02\Omega$ for $r_{\text{T}}$ in the above equation.

$\begin{array}{c}{R}_1=\left[\left(\frac{1}{12.0\Omega }\right)+\left(\frac{1}{12.0\Omega }\right)\right]+5.02\Omega \\ =6.00\Omega +5.02\Omega \\ =11.02\Omega \\ \approx 11.0\Omega \end{array}$

Thus, the total resistance of the resistor is $11.0\Omega$.

Formula to calculate the current in the batteries is,

$I_{\text{2}}=\frac{V_{\text{T}}}{R_1}$

Here,

$I_2$ is the power delivered to the resistor.

Substitute $18.6\text{V}$  for $V_{\text{T}}$, $11.0\Omega$ for $R_1$ in the above equation.

$\begin{array}{c}{I}_{\text{2}}=\frac{18.6\text{V}}{11.0\Omega }\\ =1.69\text{A}\end{array}$

Thus, the current produced in the batteries is $1.69\text{A}$.

Formula to calculate the terminal voltage across both batteries is,

$\Delta V\text{'}={\epsilon }_{\text{T}}-I\left(r_{\text{T}}\right)$

Here,

$\Delta V\text{'}$ is the terminal voltage across both batteries.

Substitute $18.6\text{V}$ for ${\epsilon }_{\text{T}}$, $1.69\text{A}$ for $I_2$, $5.02\Omega$ for $r_{\text{T}}$ in the above equation.

$\begin{array}{c}\Delta V\text{'}=18.6\text{V}-1.69\text{A}\left(5.02\Omega \right)\\ =10.11\text{V}\\ \approx \text{10}\text{.1}\text{V}\end{array}$

Thus, the terminal voltage across both batteries is $\text{10}\text{.1}\text{V}$.

Formula to calculate the power delivered to each resistor is,

$P\text{'}=\frac{\Delta V\text{'}}{R}$

Here,

$P\text{'}$ is the power delivered to each resistor.

Substitute $\text{10}\text{.1}\text{V}$ for $\Delta V\text{'}$, $12.0\Omega$ for $R$ in the above equation.

$\begin{array}{c}P\text{'}=\frac{{\left(\text{10}\text{.1}\text{V}\right)}^2}{12.0\Omega }\\ =8.50\text{W}\\ \approx \text{8}\text{.54}\text{W}\end{array}$

Conclusion:

Therefore, the power delivered to each resistor is $8.54\text{W}$.

To determine

The reason for the power in part (g) is not same as in part (f).

Answer to Problem 45AP

The internal resistance of the batteries and the terminal voltage of the batteries is not same in both cases.

Explanation of Solution

Given info: The open circuit voltage of the battery is $9.30\text{V}$. The short circuit current is $3.70\text{A}$.

In part (g), the total internal resistance of the resistor is $11.0\Omega$ and terminal voltage across both batteries is $\text{10}\text{.1}\text{V}$. In part (f), the internal resistance is $12.0\Omega$. Thus, the power is dependent upon the current and resistance. So, if the resistance is change the power is also changed.

Conclusion:

Therefore, the internal resistance of the batteries and the terminal voltage of the batteries is not same in both cases.