Chapter 12, Problem 23P

To determine

Find the current in each branch in the circuit.

Answer to Problem 23P

The current in the branch $R_1$ is $1.3\text{A}$.

The current in the branch $R_2$ is $0.73\text{A}$

The current in the branch $R_3$ is $0.57\text{A}$.

Explanation of Solution

Given data:

The value of the resistance $R_1$ is $3\text{Ω}$.

The value of the resistance $R_2$ is $7\text{Ω}$.

The value of the resistance $R_3$ is $9\text{Ω}$.

The supply voltage in the circuit is $9\text{V}$.

Formula used:

Formula to calculate the equivalent resistance in the circuit is,

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

Here,

$R_2$ is the resistance value of the resistor 2.

$R_3$ is the resistance value of the resistor 3.

Formula to calculate the total resistance in the circuit is,

$R_{\text{total}}=R_1+R_{\text{equivalent}}$ (2)

Here,

$R_{\text{equivalent}}$ is the equivalent resistance.

$R_1$ is the resistance value of the resistor 1.

By using ohm’s law to find the voltage V is,

$V=I_{\text{total}}{R}_{\text{total}}$

Here,

$I_{\text{total}}$ is the total current.

Rearrange the equation for the total current as,

$I_{\text{total}}=\frac{V}{R_{\text{total}}}$ (3)

By using current division rule to find the unknown current $I_2$,

$\begin{array}{c}{I}_2=I_{\text{total}}\left(\frac{\text{opposite}\text{resistance}}{\text{Total}\text{resistance}}\right)\\ =I_{\text{total}}\left(\frac{R_3}{R_2+R_3}\right)\end{array}$ (4)

By using current division rule to find the unknown current $I_3$,

$\begin{array}{c}{I}_3=I_{\text{total}}\times \left(\frac{\text{opposite}\text{resistance}}{\text{Total}\text{resistance}}\right)\\ =I_{\text{total}}\times \left(\frac{R_2}{R_2+R_3}\right)\end{array}$ (5)

Calculation:

Refer to Figure 12.16 in textbook for the example problem 12.6, redraw the circuit by replacing the light bulbs with resistors as shown in Figure 1, the $3\text{Ω}$ resistor is connected in series with a parallel combination of $7\text{Ω}$ and $9\text{Ω}$ resistors.

Substitute $7\text{Ω}$ for $R_2$ and $9\text{Ω}$ for $R_3$ in equation (1) to find $R_{\text{equivalent}}$,

$\begin{array}{c}\frac{1}{R_{\text{equivalent}}}=\left(\frac{1}{7\text{Ω}}\right)+\left(\frac{1}{9\text{Ω}}\right)\\ =0.2538\text{Ω}\\ R_{\text{equivalent}}=\frac{1}{0.2538\text{Ω}}\\ =3.9\text{Ω}\end{array}$

Substitute $3.9\text{Ω}$ for $R_{\text{equivalent}}$ and $3\text{Ω}$ for $R_1$ in equation (2) to find $R_{\text{total}}$,

$\begin{array}{c}{R}_{\text{total}}=\left(3\text{Ω}\right)\text{+}\left(\text{3}\text{.9}\text{Ω}\right)\\ =6.9\text{Ω}\end{array}$

Substitute $9\text{V}$ for V and $6.9\text{Ω}$ for $R_{\text{total}}$ in equation (3) to find the current $I_{\text{total}}$,

$\begin{array}{c}{I}_{\text{total}}=\frac{9\text{V}}{6.9\text{Ω}}\\ =1.3\text{A}\end{array}$

Therefore, the total current in the circuit is $1.3\text{A}$.

The current through the resistor $R_1$ is,

$I_{\text{total}}=I_1=1.3\text{A}$.

Substitute $1.3\text{A}$ for $I_{\text{total}}$, $7\text{Ω}$ for $R_2$ and $9\text{Ω}$ for $R_3$ in equation (4) to find the unknown current $I_2$,

$\begin{array}{c}{I}_2=\left(1.3\text{A}\right)\left(\frac{9\text{Ω}}{7\text{Ω}+9\text{Ω}}\right)\\ =\left(1.3\text{A}\right)\left(0.5625\text{Ω}\right)\\ I_2=0.73\text{A}\end{array}$

Substitute $1.3\text{A}$ for $I_{\text{total}}$, $7\text{Ω}$ for $R_2$ and $9\text{Ω}$ for $R_3$ in equation (5) to find the unknown current $I_3$,

$\begin{array}{c}{I}_3=\left(1.3\text{A}\right)\left(\frac{7\text{Ω}}{7\text{Ω}+9\text{Ω}}\right)\\ =\left(1.3\text{A}\right)\left(0.4375\text{Ω}\right)\\ I_3=0.57\text{A}\end{array}$

Conclusion:

Thus,

The current in the branch $R_1$ is $1.3\text{A}$.

The current in the branch $R_2$ is $0.73\text{A}$

The current in the branch $R_3$ is $0.57\text{A}$.