Chapter 18, Problem 10P

Consider the circuit shown in Figure P18.10. (a) Calculate the equivalent resistance of the 10.0-Ω and 5.00-Ω resistors connected in parallel. (b) Using the result of part (a), calculate the combined resistance of the 10.0-Ω, 5.00-Ω, and 4.00-Ω resistors. (c) Calculate the equivalent resistance of the combined resistance found in part (b) and the parallel 3.00-Ω resistor. (d) Combine the equivalent resistance found in part (c) with the 2.00-Ω resistor. (e) Calculate the total current in the circuit. (f) What is the voltage drop across the 2.00-Ω resistor? (g) Subtracting the result of part (f) from the battery voltage, find the voltage across the 3.00-Ω resistor. (h) Calculate the current in the 3.00-Ω resistor.

Figure P18.10

To determine

The equivalent resistance of the first parallel combination of resistors.

Answer to Problem 10P

The equivalent resistance of parallel combination resistor is $3.33\Omega$.

Explanation of Solution

Given Info: The resistors connected in parallel combination is $10.0\Omega$ and $5.00\Omega$.

Explanation:

Formula to calculate the equivalent resistance of the first parallel combination resistors is,

$\frac{1}{R_{p1}}=\frac{1}{R_1}+\frac{1}{R_2}$

• $R_{p1}$ is the equivalent resistance of parallel combination resistor,
• $R_1$ and $R_2$ is the resistors,

Substitute $10.0\Omega$ for $R_1$ and $5.00\Omega$ for $R_2$.

$\begin{array}{c}\frac{1}{R_{p1}}=\frac{1}{10.0\Omega }+\frac{1}{5.00\Omega }\\ =\frac{3}{10.0\Omega }\\ R_{p1}=\frac{10.0\Omega }{3}\\ =3.33\Omega \end{array}$

Conclusion:

Therefore, the equivalent resistance of parallel combination resistor is $3.33\Omega$.

To determine

The combined resistance of the resistors in series combination using the result in part (a).

Answer to Problem 10P

The combined resistance of the resistors in series combination is $7.33\Omega$.

Explanation of Solution

Given Info: The resistors connected in series combination is $4.00\Omega$ and equivalent resistance of parallel combination is $3.33\Omega$.

Explanation:

Formula to calculate the combined resistance of the resistors in series combination is,

$R_{\text{upper}}=R_{p1}+R_3$

• $R_{p1}$ is the equivalent resistance of parallel combination resistor,
• $R_3$ is the resistors,
• $R_{\text{upper}}$ is the combination of upper resistors mentioned in diagram,

Substitute $3.33\Omega$ for $R_{p1}$ and $4.00\Omega$ for $R_3$.

$\begin{array}{c}{R}_{\text{upper}}=3.33\Omega +4.00\Omega \\ =7.33\Omega \end{array}$

Conclusion:

Therefore, the combined resistance of the resistors in series combination is $7.33\Omega$.

To determine

The equivalent resistance of the second parallel combination of resistors shown in the diagram.

Answer to Problem 10P

The equivalent resistance of second parallel combination resistor is $2.13\Omega$.

Explanation of Solution

Given Info: The combined resistance of the resistors is $7.33\Omega$ and the parallel resistor in the given circuit is $3.00\Omega$.

Explanation:

Formula to calculate the equivalent resistance of the second parallel combination resistors is,

$\frac{1}{R_{p2}}=\frac{1}{R_{\text{upper}}}+\frac{1}{R_4}$

• $R_{p2}$ is the equivalent resistance of parallel combination resistor,
• $R_4$ is the resistor,

Rearrange the above relation as,

$R_{p2}=\frac{R_{\text{upper}}{R}_4}{R_{\text{upper}}+R_4}$

Substitute $7.33\Omega$ for $R_{\text{upper}}$ and $3.00\Omega$ for $R_4$.

$\begin{array}{c}{R}_{p2}=\frac{\left(7.33\Omega \right)\left(3.00\Omega \right)}{7.33\Omega +3.00\Omega }\\ =2.13\Omega \end{array}$

Conclusion:

Therefore, the equivalent resistance of second parallel combination resistor is $2.13\Omega$.

To determine

The combined resistance of the resistors in second series combination for the entire circuit.

Answer to Problem 10P

The combined resistance of the resistors in second series combination is $4.13\Omega$.

Explanation of Solution

Given Info: The resistors connected in series combination is $2.00\Omega$ and equivalent resistance of second parallel combination is $2.13\Omega$.

Explanation:

Formula to calculate the combined resistance of the resistors in second series combination is,

$R_{\text{total}}=R_{p2}+R_5$

• $R_{p2}$ is the equivalent resistance of second parallel combination resistor,
• $R_5$ is the resistors,
• $R_{\text{total}}$ is the total resistance in entire circuit,

Substitute $2.00\Omega$ for $R_5$ and $2.13\Omega$ for $R_{p2}$.

$\begin{array}{c}{R}_{\text{total}}=2.13\Omega +2.00\Omega \\ =4.13\Omega \end{array}$

Conclusion:

Therefore, the combined resistance of the resistors in second series combination is $4.13\Omega$.

To determine

The total current passing through the circuit.

Answer to Problem 10P

The total current passing through the circuit is 1.94 A.

Explanation of Solution

Given Info: The total resistance is $4.13\Omega$ and potential difference across the battery is $8.00\text{V}$.

Explanation:

Formula to calculate the total current in the circuit is,

$I_{\text{total}}=\frac{\Delta V}{R_{\text{total}}}$

• $I_{\text{total}}$ is the total current,
• $\Delta V$ is the potential difference,
• $R_{\text{total}}$ is the total resistance,

Substitute $8.00\text{V}$ for $\Delta V$ and $4.13\Omega$ for $R_{\text{total}}$.

$\begin{array}{c}{I}_{\text{total}}=\frac{8.00\text{V}}{4.13\Omega }\\ =1.94\text{A}\end{array}$

Conclusion:

Therefore, the total current passing through the circuit is 1.94 A.

To determine

The potential drop across the $2.00\Omega$ resistor.

Answer to Problem 10P

The potential drop across the $2.00\Omega$ resistor is 3.88 V.

Explanation of Solution

Given Info: The total current in the circuit is 1.94 A and resistor is $2.00\Omega$.

Explanation:

Formula to calculate the potential drop is,

$\Delta V_2=R_5{I}_{total}$

• $I_{\text{total}}$ is the total current,
• $\Delta V_2$ is the potential drop across the $2.00\Omega$ resistor,
• $R_5$ is the resistor,

Substitute $1.94\text{A}$ for $I_{\text{total}}$ and $2.00\Omega$ for $R_5$.

$\begin{array}{c}\Delta V_2=\left(2.00\Omega \right)\left(1.94\text{A}\right)\\ =3.88\text{V}\end{array}$

Conclusion:

Therefore, the potential drop across the $2.00\Omega$ resistor is 3.88 V.

To determine

The potential drop across the second parallel combination resistance.

Answer to Problem 10P

The potential drop across the second parallel combination is 4.12 V.

Explanation of Solution

Given Info: The potential difference across the battery is $8.00\text{V}$ and the potential drop across the $2.00\Omega$ resistor is $3.88\text{V}$.

Explanation:

Formula to calculate the potential drop across the second parallel combination is,

$\Delta V_{p2}=\Delta V-\Delta V_2$

• $\Delta V_{p2}$ is the potential drop across the second parallel combination,
• $\Delta V$ is the potential difference across the battery,

Substitute $3.88\text{V}$ for $\Delta V_2$ and $8.00\text{V}$ for $\Delta V$.

$\begin{array}{c}\Delta V_{p2}=8.00\text{V}-3.88\text{V}\\ =4.12\text{V}\end{array}$

Conclusion:

Therefore, the potential drop across the second parallel combination is 4.12 V.

To determine

The current passing through the $3.00\Omega$ resistor in the circuit.

Answer to Problem 10P

The total current passing through the $3.00\Omega$ resistor is 1.37 A.

Explanation of Solution

Given Info: The potential drop across the second parallel combination is 4.12 V. and resistor is $3.00\Omega$.

Explanation:

Formula to calculate the current passing through the $3.00\Omega$ resistor is,

$I_{\text{total}}=\frac{\Delta V_{p2}}{R_4}$

• $I_{\text{total}}$ is the total current,
• $R_4$ is the resistor,

Substitute $4.12\text{V}$ for $\Delta V_{p2}$ and $3.00\Omega$ for $R_4$.

$\begin{array}{c}{I}_{\text{total}}=\frac{4.12\text{V}}{3.00\Omega }\\ =1.37\text{A}\end{array}$

Conclusion:

Therefore, the total current passing through the $3.00\Omega$ resistor is 1.37 A.