Chapter 18, Problem 61AP

A battery with an internal resistance of 10.0 Ω produces an open circuit voltage of 12.0 V. A variable load resistance with a range from 0 to 30.0 Ω is connected across the battery. (Note: A battery has a resistance that depends on the condition of its chemicals and that increases as the battery ages. This internal resistance can be represented in a simple circuit diagram as a resistor in series with the battery.) (a) Graph the power dissipated in the load resistor as a function of the load resistance. (b) With your graph, demonstrate the following important theorem: The power delivered to a load is a maximum if the load resistance equals the internal resistance of the source.

To determine

The equivalent resistance and current in each resistor when the power is connected between points A and B.

Answer to Problem 61AP

The equivalent resistance of the circuit is $0.099\Omega$.

The current through the resistor $R_1$ is $50.0\text{A}$ and the current through the resistors $R_2$, $R_3$ and $100\Omega$ is $0.0450\text{A}$.

Explanation of Solution

Given Info:

The power supply is $5.0\text{V}$.

The resistance $R_1$ is $0.100\Omega$.

The resistance $R_2$ is $1.0\Omega$.

The resistance $R_3$ is $10.0\Omega$.

The total resistance of $R_2$, $R_3$ and $100\Omega$ in series connection is $111.0\Omega$.

The power supply is $5.0\text{V}$.

Explanation:

The equivalent resistance is,

When the power supply is connected between A and B, the resistors $R_2$, $R_3$ and $100\Omega$ are in series. The total resistance of $R_2$, $R_3$ and $100\Omega$ are in parallel with the $R_1$ resistance.

Thus, the equivalent resistance of the circuit is,

$\frac{1}{R_{equ}}=\frac{1}{R_1}+\frac{1}{R_2+R_3+100\Omega }$

• $R_1$ is the first resistance
• $R_2$ is the second resistance
• $R_3$ is the third resistance

Substitute $0.100\Omega$ for $R_1$, $1.0\Omega$ for $R_2$ and $10.0\Omega$ for $R_3$ to find the equivalent resistance.

$\begin{array}{c}{R}_{equ}=\frac{1}{\frac{1}{0.100\Omega }+\frac{1}{1.0\Omega +10.0\Omega +100\Omega }}\\ =\frac{1}{\frac{1}{0.100\Omega }+\frac{1}{111\Omega }}\\ =0.099\Omega \end{array}$

The total resistance of $R_2$, $R_3$ and $100\Omega$ is $111.0\Omega$.

Thus, the equivalent resistance of the circuit is $0.099\Omega$.

Formula to calculate the current is,

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

• V is the voltage across the terminal
• R is the equivalent resistance of the circuit

For resistor $R_1$,

Substitute $5.0\text{V}$ for $V$ and $0.100\Omega$ for $R_1$ to find the current through the resistor $R_1$.

$\begin{array}{c}{I}_{R_1}=\frac{5.0\text{V}}{0.100\Omega }\\ =50.0\text{A}\end{array}$

Since, the resistors $R_2$, $R_3$ and $100\Omega$ are in a series connection; the current in these resistors will be the same.

For $R_2$, $R_3$ and $100\Omega$,

Substitute $5.0\text{V}$ for $V$ and $111.0\Omega$ for $R$ to find the current through the resistors $R_2$, $R_3$ and $100\Omega$,

$\begin{array}{c}{I}_{R_2}=I_{R_3}=I_{R_{100}}=\frac{5.0\text{V}}{111.0\Omega }\\ =0.0450\text{A}\end{array}$

Thus, the current through the resistor $R_1$ is $50.0\text{A}$ and the current through the resistors $R_2$, $R_3$ and $100\Omega$ is $0.0450\text{A}$.

Conclusion:

The equivalent resistance of the circuit is $0.099\Omega$.

The current through the resistor $R_1$ is $50.0\text{A}$ and the current through the resistors $R_2$, $R_3$ and $100\Omega$ is $0.0450\text{A}$.

To determine

The equivalent resistance and current in each resistor when the power is connected between points A and C.

Answer to Problem 61AP

The equivalent resistance of the circuit is $1.09\Omega$.

The current through the resistor $R_1$ and $R_2$ is $4.55\text{A}$ and the current through the resistors $R_3$ and $100\Omega$ is $0.0455\text{A}$.

Explanation of Solution

Given Info:

The power supply is $5.0\text{V}$.

The resistance $R_1$ is $0.100\Omega$.

The resistance $R_2$ is $1.0\Omega$.

The resistance $R_3$ is $10.0\Omega$.

The total resistance of $R_2$, $R_1$ is $1.10\Omega$.

The total resistance of $R_3$, $R_{100}$ is $110.0\Omega$.

The power supply is $5.0\text{V}$.

Explanation:

The equivalent resistance is,

When the power supply is connected between A and C, the resistors $R_1$, $R_2$ are in series connection and the resistors $R_3$ and. $100.0\Omega$ are in series connection. The total resistance of $R_2$, $R_1$ is in parallel with the total resistance of $R_1$ and $100\Omega$ resistance.

Thus, the equivalent resistance of the circuit is,

$\frac{1}{R_{equ}}=\frac{1}{R_1+R_2}+\frac{1}{R_3+100\Omega }$

• $R_1$ is the first resistance
• $R_2$ is the second resistance
• $R_3$ is the third resistance

Substitute $0.100\Omega$ for $R_1$, $1.0\Omega$ for $R_2$ and $10.0\Omega$ for $R_3$ to find the equivalent resistance.

$\begin{array}{c}{R}_{equ}=\frac{1}{\frac{1}{0.100\Omega +1.0\Omega }+\frac{1}{10.0\Omega +100\Omega }}\\ =\frac{1}{\frac{1}{1.10\Omega }+\frac{1}{110\Omega }}\\ =1.09\Omega \end{array}$

The total resistance of $R_2$, $R_1$ is $1.10\Omega$.

The total resistance of $R_3$, $R_{100}$ is $110.0\Omega$.

Thus, the equivalent resistance of the circuit is $1.09\Omega$.

Formula to calculate the current is,

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

• V is the voltage across the terminal
• R is the equivalent resistance of the circuit

For resistor $R_1$ and $R_2$,

Substitute $5.0\text{V}$ for $V$ and $1.10\Omega$ for $R$ to find the current through the resistors $R_1$ and $R_2$,

$\begin{array}{c}{I}_{R_1}=I_{R_2}=\frac{5.0\text{V}}{1.10\Omega }\\ =4.55\text{A}\end{array}$

Since, the resistors $R_3$ and $100\Omega$ are in a series connection; the current in these resistors will be the same.

For $R_3$ and $100\Omega$,

Substitute $5.0\text{V}$ for $V$ and $110.0\Omega$ for $R$ to find the current through the resistors $R_3$ and $100\Omega$,

$\begin{array}{c}{I}_{R_3}=I_{R_{100}}=\frac{5.0\text{V}}{110.0\Omega }\\ =0.0455\text{A}\end{array}$

Thus, the current through the resistor $R_1$ and $R_2$ is $4.55\text{A}$ and the current through the resistors $R_3$ and $100\Omega$ is $0.0455\text{A}$.

Conclusion:

The equivalent resistance of the circuit is $1.09\Omega$.

The current through the resistor $R_1$ and $R_2$ is $4.55\text{A}$ and the current through the resistors $R_3$ and $100\Omega$ is $0.0455\text{A}$.

To determine

The equivalent resistance and current in each resistor when the power is connected between points A and D.

Answer to Problem 61AP

The equivalent resistance of the circuit is $9.99\Omega$.

The current through the resistor $R_3$, $R_1$ and $R_2$ is $0.450\text{A}$ and the current through the resistor $100\Omega$ is $0.050\text{A}$.

Explanation of Solution

Given Info:

The power supply is $5.0\text{V}$.

The resistance $R_1$ is $0.100\Omega$.

The resistance $R_2$ is $1.0\Omega$.

The resistance $R_3$ is $10.0\Omega$.

The total resistance of $R_2$, $R_1$ and $R_3$ is $11.1\Omega$.

The power supply is $5.0\text{V}$.

Explanation:

The equivalent resistance is,

When the power supply is connected between A and D, the resistors $R_1$, $R_2$ and $R_3$ are in series connection. The total resistance of $R_2$, $R_1$ and $R_3$ is in parallel with the resistance $100\Omega$.

Thus, the equivalent resistance of the circuit is,

$\frac{1}{R_{equ}}=\frac{1}{R_1+R_2+R_3}+\frac{1}{100\Omega }$

• $R_1$ is the first resistance
• $R_2$ is the second resistance
• $R_3$ is the third resistance

Substitute $0.100\Omega$ for $R_1$, $1.0\Omega$ for $R_2$ and $10.0\Omega$ for $R_3$ to find the equivalent resistance.

$\begin{array}{c}{R}_{equ}=\frac{1}{\frac{1}{0.100\Omega +1.0\Omega +10.0\Omega }+\frac{1}{100\Omega }}\\ =\frac{1}{\frac{1}{11.1\Omega }+\frac{1}{100\Omega }}\\ =9.99\Omega \end{array}$

The total resistance of $R_2$, $R_1$ and $R_3$ is $11.1\Omega$.

Thus, the equivalent resistance of the circuit is $9.99\Omega$.

Formula to calculate the current is,

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

• V is the voltage across the terminal
• R is the equivalent resistance of the circuit

For resistor $R_3$, $R_1$ and $R_2$,

Substitute $5.0\text{V}$ for $V$ and $11.1\Omega$ for $R$ to find the current through the resistors $R_3$, $R_1$ and $R_2$,

$\begin{array}{c}{I}_{R_1}=I_{R_2}=I_{R_3}=\frac{5.0\text{V}}{11.1\Omega }\\ =0.450\text{A}\end{array}$

For $100\Omega$,

Substitute $5.0\text{V}$ for $V$ and $100.0\Omega$ for $R$ to find the current through the resistor $100\Omega$,

$\begin{array}{c}{I}_{R_{100}}=\frac{5.0\text{V}}{100.0\Omega }\\ =0.050\text{A}\end{array}$

Thus, the current through the resistor $R_3$, $R_1$ and $R_2$ is $0.450\text{A}$ and the current through the resistor $100\Omega$ is $0.050\text{A}$.

Conclusion:

The equivalent resistance of the circuit is $9.99\Omega$.

The current through the resistor $R_3$, $R_1$ and $R_2$ is $0.450\text{A}$ and the current through the resistor $100\Omega$ is $0.050\text{A}$.