Chapter 15, Problem 8SA

Solve the parallel circuit for total circuit resistance, total circuit amperage, and the amperage through each leg (Figure 15-65).

To determine

The total circuit resistance.

Total circuit amperage.

The amperage through leg 1.

The amperage through leg 2.

The amperage through leg 3.

Answer to Problem 8SA

The total circuit resistance is $1\Omega$.

Total circuit amperage is $12\text{A}$.

The amperage through leg 1 is $2\text{A}$.

The amperage through leg2 is $4\text{A}$.

The amperage through leg3 is $6\text{A}$.

Explanation of Solution

Given information:

The total voltage is $12\text{V}$, the first resistance is $6\Omega$, the second resistance is $3\Omega$ and the third resistance is $2\Omega$.

Write the expression of total resistance of the circuit.

  $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$  ....... (I)

Here, the third resistance is $R_3$, the first resistance is $R_1$ and the second resistance is $R_2$.

Write the expression of total current or amperage.

  $I=\frac{V}{R}$  ....... (II)

Here, the total voltage is $V$ and the total resistance is $R$.

Write the expression of current or amperage through leg 1.

  $I_1=\frac{V}{R_1}$  ....... (III)

Write the expression of current or amperage through leg2.

  $I_2=\frac{V}{R_2}$ ....... (IV)

Write the expression of current or amperage through leg 3.

  $I_3=\frac{V}{R_3}$  ....... (V)

Calculation:

Substitute $6\Omega$ for $R_1$, $3\Omega$ for $R_2$ and $2\Omega$ for $R_3$ in Equation (I).

  $\begin{array}{l}\frac{1}{R}=\frac{1}{6\Omega }+\frac{1}{3\Omega }+\frac{1}{2\Omega }\\ \frac{1}{R}=1{\Omega }^{-1}\\ R=1\Omega \end{array}$

Thus, the total circuit resistance is $1\Omega$.

Substitute $12\text{V}$ for $V$ and $1\Omega$ for $R$ in Equation (II).

  $\begin{array}{c}I=\frac{12\text{V}}{1\Omega }\\ =\left(12\text{V}/\Omega \right)\times \left(\frac{1\text{A}}{1\text{V}/\Omega }\right)\\ =12\text{A}\end{array}$

Thus, total circuit amperage is $12\text{A}$.

Substitute $12\text{V}$ for $V$ and $6\Omega$ for $R_1$ in Equation (III).

  $\begin{array}{c}{I}_1=\frac{12\text{V}}{6\Omega }\\ =\left(2\text{V}/\Omega \right)\times \left(\frac{1\text{A}}{1\text{V}/\Omega }\right)\\ =2\text{A}\end{array}$

Thus, the amperage through leg 1 is $2\text{A}$.

Substitute $12\text{V}$ for $V$ and $3\Omega$ for $R_2$ in Equation (IV).

  $\begin{array}{c}{I}_2=\frac{12\text{V}}{3\Omega }\\ =\left(4\text{V}/\Omega \right)\times \left(\frac{1\text{A}}{1\text{V}/\Omega }\right)\\ =4\text{A}\end{array}$

Thus, the amperage through leg2 is $4\text{A}$.

Substitute $12\text{V}$ for $V$ and $2\Omega$ for $R_3$ in Equation (V).

  $\begin{array}{c}{I}_3=\frac{12\text{V}}{2\Omega }\\ =\left(6\text{V}/\Omega \right)\times \left(\frac{1\text{A}}{1\text{V}/\Omega }\right)\\ =6\text{A}\end{array}$

Thus, the amperage through leg 3 is $6\text{A}$.