Chapter 13, Problem 1SP

In the circuit shown, the internal resistance of the battery can be considered negligible.

a.    What is the equivalent resistance of the two-resistor parallel combination?

b.    What is the total current flowing through the battery?

c.     What is the voltage drop across the 12-ohm resistor?

d.     What is the voltage drop across the 15-ohm resistor?

e.    What is the current flowing through the 15 Ω resistor?

f.    What is the power dissipated in the 12 Ω resistor?

g.    Is the current flowing through the 12 Ω resistor greater or less than that flowing through the 15 Ω resistor? Explain.

To determine

The equivalent resistance of the two parallel resistors.

Answer to Problem 1SP

The equivalent resistance of the two parallel resistors is $10\text{Ω}$.

Explanation of Solution

Given info: The given circuit is shown below.

In the given circuit the resistances $15\text{Ω}$ and $30\text{Ω}$ are in parallel. The resistance $12\text{Ω}$ is in series with the parallel connection of $15\text{Ω}$ and $30\text{Ω}$.

Write the formula for the equivalent resistance of two resistances connected in parallel.

$R_p=\frac{1}{\left(\frac{1}{R_1}+\frac{1}{R_2}\right)}$

Here,

$R_p$ is the equivalent resistance of the two parallel resistances

$R_1$ is the first resistance

$R_2$ is the second resistance

Substitute $15\text{Ω}$ for $R_1$, $30\text{Ω}$ for $R_2$ to determine $R_p$.

$\begin{array}{c}{R}_p=\frac{1}{\left(\frac{1}{15\text{Ω}}+\frac{1}{30\text{Ω}}\right)}\\ =10\text{Ω}\end{array}$

Conclusion:

The equivalent resistance of the two parallel resistors is $10\text{Ω}$.

To determine

The total current flowing in the circuit.

Answer to Problem 1SP

The total current through the circuit is $0.273\text{A}$.

Explanation of Solution

Given info: The given circuit is shown below.

In the given circuit the resistances $15\text{Ω}$ and $30\text{Ω}$ are in parallel. The resistance $12\text{Ω}$ is in series with the parallel connection of $15\text{Ω}$ and $30\text{Ω}$.

Write the formula for the total resistance of the circuit.

$R=R_p+R_3$

Here,

$R_p$ is the equivalent resistance of the two parallel resistances

$R$ is the total resistance of the circuit

$R_3$ is the third resistance

Write the formula for the total current in the circuit.

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

Here,

$I$ is the current

$V$ is the voltage of the battery

Substitute the expression of $R$ in the above expression.

$I=\frac{V}{R_p+R_3}$

From section (a) the equivalent resistance of the two parallel resistances is $10\text{Ω}$.

Substitute $12\text{Ω}$ for $R_3$, $10\text{Ω}$ for $R_p$, $6.0\text{V}$ for $V$ to determine $I$.

$\begin{array}{c}I=\frac{6.0\text{V}}{10\text{Ω}+12\text{Ω}}\\ =0.273\text{A}\end{array}$

Conclusion:

The total current through the circuit is $0.273\text{A}$.

To determine

The voltage drop across $12\text{Ω}$.

Answer to Problem 1SP

The voltage drop across $12\text{Ω}$ is $3.27\text{V}$.

Explanation of Solution

Given info: The given circuit is shown below.

In the given circuit the resistances $15\text{Ω}$ and $30\text{Ω}$ are in parallel. The resistance $12\text{Ω}$ is in series with the parallel connection of $15\text{Ω}$ and $30\text{Ω}$.

The voltage of the battery will get distributed among the $12\text{Ω}$ and the parallel combination of $15\text{Ω}$ and $30\text{Ω}$.

Write the formula for the voltage across the $R_3$.

$V_3=\left(\frac{R_3}{R_p+R_3}\right)V$

Here,

$V_3$ is the voltage across $R_3$

$R_3$ is the third resistance

$R_p$ is the equivalent resistance of the two parallel resistors

$V$ is the voltage of the battery

From section (a) the equivalent resistance of the parallel connection is $10\text{Ω}$.

Substitute $10\text{Ω}$ for $R_p$, $12\text{Ω}$ for $R_3$, $6.0\text{V}$ for $V$ to determine $V_3$.

$\begin{array}{c}{V}_3=\left(\frac{12\text{Ω}}{10\text{Ω}+12\text{Ω}}\right)\left(6.0\text{V}\right)\\ =3.27\text{V}\end{array}$

Conclusion:

The voltage drop across $12\text{Ω}$ is $3.27\text{V}$.

To determine

The voltage drop across $15\text{Ω}$.

Answer to Problem 1SP

The voltage drop across $15\text{Ω}$ is $\text{2}\text{.73}\text{V}$.

Explanation of Solution

Given info: The given circuit is shown below.

In the given circuit the resistances $15\text{Ω}$ and $30\text{Ω}$ are in parallel. The resistance $12\text{Ω}$ is in series with the parallel connection of $15\text{Ω}$ and $30\text{Ω}$.

The voltage of the battery will get distributed among the $12\text{Ω}$ and the parallel combination of $15\text{Ω}$ and $30\text{Ω}$.

The voltage drop across $15\text{Ω}$ and $30\text{Ω}$ are equal.

Write the formula for the voltage across the parallel connection of $15\text{Ω}$ and $30\text{Ω}$.

$V_p=V-V_s$

Here,

$V_p$ is the voltage across the parallel resistors

$V$ is the total voltage of the battery

$V_s$ is the voltage of the series resistor

From section (c) the voltage across the series resistor is $3.27\text{V}$.

Substitute $3.27\text{V}$ for $V_s$, $6.0\text{V}$ for $V$ to determine $V_p$.

$\begin{array}{c}{V}_p=6.0\text{V}-3.27\text{V}\\ \text{=2}\text{.73}\text{V}\end{array}$

Conclusion:

The voltage drop across $15\text{Ω}$ is $\text{2}\text{.73}\text{V}$.

To determine

The current flowing through $15\text{Ω}$.

Answer to Problem 1SP

The current flowing through $15\text{Ω}$ is $0.182\text{A}$.

Explanation of Solution

Given info: The given circuit is shown below.

In the given circuit the resistances $15\text{Ω}$ and $30\text{Ω}$ are in parallel. The resistance $12\text{Ω}$ is in series with the parallel connection of $15\text{Ω}$ and $30\text{Ω}$.

Write the formula for current.

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

Here,

$I$ is the current

$V$ is the voltage

$R$ is the resistance

The voltage drop across $15\text{Ω}$ is $\text{2}\text{.73}\text{V}$.

Substitute $\text{2}\text{.73}\text{V}$ for $V$, $15\text{Ω}$ for $R$ to determine $I$.

$\begin{array}{c}I=\frac{\text{2}\text{.73}\text{V}}{15\text{Ω}}\\ =0.182\text{A}\end{array}$

Conclusion:

The current flowing through $15\text{Ω}$ is $0.182\text{A}$.

To determine

The power dissipated in $12\text{Ω}$.

Answer to Problem 1SP

The power dissipated in $12\text{Ω}$ is $0.891\text{W}$.

Explanation of Solution

Given info: The given circuit is shown below.

In the given circuit the resistances $15\text{Ω}$ and $30\text{Ω}$ are in parallel. The resistance $12\text{Ω}$ is in series with the parallel connection of $15\text{Ω}$ and $30\text{Ω}$.

Write the formula for power.

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

Here,

$R$ is the resistance

$V$ is the voltage

$P$ is the power

From section (c) the voltage drop across $12\text{Ω}$ is $3.27\text{V}$.

Substitute $3.27\text{V}$ for $V$, $12\text{Ω}$ for $R$ to determine $I$.

$\begin{array}{c}P=\frac{{\left(3.27\text{V}\right)}^2}{12\text{Ω}}\\ =0.891\text{W}\end{array}$

Conclusion:

The power dissipated in $12\text{Ω}$ is $0.891\text{W}$.

To determine

Whether the current through $12\text{Ω}$ is greater or lesser than the current through $15\text{Ω}$.

Answer to Problem 1SP

The current through $12\text{Ω}$ is greater than the current through $15\text{Ω}$.

Explanation of Solution

Given info: The given circuit is shown below.

In the given circuit the resistances $15\text{Ω}$ and $30\text{Ω}$ are in parallel. The resistance $12\text{Ω}$ is in series with the parallel connection of $15\text{Ω}$ and $30\text{Ω}$.

Since the current is a conserved quantity, in the given circuit, the current through $12\text{Ω}$ is equal to the sum of current through $15\text{Ω}$ and $30\text{Ω}$. Thus the current through $12\text{Ω}$ is greater than the current through $15\text{Ω}$.

Conclusion:

The current through $12\text{Ω}$ is greater than the current through $15\text{Ω}$.