Chapter 12, Problem 19P

To determine

Find the total resistance and the current flow in each branch for the circuit.

Answer to Problem 19P

The total resistance is $1.5\text{Ω}$.

The current flows through the branch $R_1$ is $2.0\text{A}$.

The current flows through the branch $R_2$ is $4.0\text{A}$.

The current flows through the branch $R_3$ is $2.0\text{A}$.

Explanation of Solution

Given data:

The supply voltage is $12\text{V}$.

The value of the resistor $R_1$ is $6\text{Ω}$.

The value of the resistor $R_2$ is $3\text{Ω}$.

The value of the resistor $R_3$ is $6\text{Ω}$.

Formula used:

Formula to calculate the total resistance in a parallel circuit,

$\frac{1}{R_{\text{total}}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ (1)

Here,

$R_1$, $R_2$, $R_3$ are the resistances.

Formula to calculate the voltage across the resistor $R_1$,

$V=I_1{R}_1$

Here,

$I_1$ is the current flow through the resistance $R_1$.

$R_1$ is the resistance.

Rearrange the equation for the current flow through the resistance $R_1$,

$I_1=\frac{V}{R_1}$ (2)

Formula to calculate the voltage across the resistor $R_2$,

$V=I_2{R}_2$

Here,

$R_2$ is the resistance.

$I_2$ is the current flow through the resistance $R_2$.

Rearrange the equation for the current flow through the resistance $R_2$,

$I_2=\frac{V}{R_2}$ (3)

Formula to calculate the voltage across the resistor $R_3$,

$V=I_3{R}_3$

Here,

$R_3$ is the resistance.

$I_3$ is the current flow through the resistance $R_3$.

Rearrange the equation for the current flow through the resistance $R_3$,

$I_3=\frac{V}{R_3}$ (4)

Formula to calculate the total current drawn by the circuit,

$I_{\text{total}}=I_1+I_2+I_3$ (5)

Calculation:

Refer to Figure problem 12.19 in the textbook, and redraw it as Figure 1, with the two light bulbs represents the resistors $\left(R_1,R_3\right)$ are connected in a parallel arrangement with the resistor $R_2$.

Substitute $6\text{Ω}$ for $R_1$, $3\text{Ω}$ for $R_2$ and $6\text{Ω}$ for $R_3$ in equation (1) to find $R_{\text{total}}$,

$\begin{array}{c}\frac{1}{R_{\text{total}}}=\frac{1}{6\Omega }+\frac{1}{3\Omega }+\frac{1}{6\Omega }\\ =0.662\Omega \end{array}$

Reduce the equation as,

$\begin{array}{c}{R}_{\text{total}}=\frac{1}{0.662\Omega }\\ R_{\text{total}}=1.5\Omega \end{array}$

The voltage drop across the each light bulb and the resistor is equal to the $12\text{V}$.

$\begin{array}{c}V=V_{\text{R1}}=V_{\text{R2}}=V_{\text{R3}}\\ =12\text{V}\end{array}$

Substitute $12\text{V}$ for V and $6\text{Ω}$ for $R_1$ in equation (2) to find $I_1$,

$\begin{array}{c}{I}_1=\frac{12\text{V}}{6\text{Ω}}\\ =2\text{A}\end{array}$

Substitute $12\text{V}$ for V and $3\text{Ω}$ for $R_2$ in equation (3) to find $I_2$,

$\begin{array}{c}{I}_2=\frac{12\text{V}}{3\text{Ω}}\\ =4\text{A}\end{array}$

Substitute $12\text{V}$ for V and $6\text{Ω}$ for $R_3$ in equation (4) to find $I_3$,

$\begin{array}{c}{I}_3=\frac{12\text{V}}{6\text{Ω}}\\ =2\text{A}\end{array}$

Substitute $2\text{A}$ for $I_1$, $4\text{A}$ for $I_2$ and $2\text{A}$ for $I_3$ in equation (5) to find $I_{\text{total}}$,

$\begin{array}{c}{I}_{\text{total}}=2\text{Ω+4}\text{Ω}+2\text{Ω}\\ =8\text{Ω}\end{array}$

Conclusion:

Hence,

The total resistance is $1.5\text{Ω}$.

The current flows through the branch $R_1$ is $2.0\text{A}$.

The current flows through the branch $R_2$ is $4.0\text{A}$.

The current flows through the branch $R_3$ is $2.0\text{A}$.