Question

in part (d) of problems 1, 2, and 3, you calculated the total resistance of each circuit. this required you to first find the current in each branch. then you found the total current and used ohms law to calculate the total resistance. another way to find the total resistance of two parallel resistors is to use the formula shown below.

$r_{total}=\frac{r_1\times r_2}{r_1 + r_2}$

example

calculate the total resistance of a circuit containing two 6 ohm resistors.

given
the circuit contains two 6 ω resistors in parallel.

looking for
the total resistance of the circuit.

relationships
$r_{total}=\frac{r_1\times r_2}{r_1 + r_2}$

solution
$r_{total}=\frac{6 omega\times6 omega}{6 omega + 6 omega}$
$r_{total}=3 omega$
the total resistance is 3 ohms.

practice 2

1. calculate the total resistance of a circuit containing each of the following combinations of resistors.

a. two 8 ω resistors in parallel

b. two 12 ω resistors in parallel

c. a 4 ω resistor and an 8 ω resistor in parallel

d. a 12 ω resistor and a 3 ω resistor in parallel

1. to find the total resistance of three resistors a, b, and c in parallel, first use the formula to find the total of resistors a and b. then use the formula again to combine resistor c with the total of a and b. use this method to find the total resistance of a circuit containing each of the following combinations of resistors

a. three 8 ω resistors in parallel

b. two 6 ω resistors and a 2 ω resistor in parallel

c. a 1 ω, a 2 ω, and a 3 ω resistor in parallel

Explanation:

Step1: Recall parallel - resistance formula

The formula for the total resistance $R_{total}$ of two parallel resistors is $R_{total}=\frac{R_1\times R_2}{R_1 + R_2}$.

Step2: Solve part 1a

Given $R_1 = R_2=8\Omega$, then $R_{total}=\frac{8\Omega\times8\Omega}{8\Omega + 8\Omega}=\frac{64\Omega^2}{16\Omega}=4\Omega$.

Step3: Solve part 1b

Given $R_1 = R_2 = 12\Omega$, then $R_{total}=\frac{12\Omega\times12\Omega}{12\Omega+12\Omega}=\frac{144\Omega^2}{24\Omega}=6\Omega$.

Step4: Solve part 1c

Given $R_1 = 4\Omega$ and $R_2 = 8\Omega$, then $R_{total}=\frac{4\Omega\times8\Omega}{4\Omega + 8\Omega}=\frac{32\Omega^2}{12\Omega}=\frac{8}{3}\Omega\approx2.67\Omega$.

Step5: Solve part 1d

Given $R_1 = 12\Omega$ and $R_2 = 3\Omega$, then $R_{total}=\frac{12\Omega\times3\Omega}{12\Omega+3\Omega}=\frac{36\Omega^2}{15\Omega}=\frac{12}{5}\Omega = 2.4\Omega$.

Step6: Solve part 2a

First, for two $8\Omega$ resistors, $R_{12}=\frac{8\Omega\times8\Omega}{8\Omega + 8\Omega}=4\Omega$. Then, combining with the third $8\Omega$ resistor, $R_{total}=\frac{4\Omega\times8\Omega}{4\Omega + 8\Omega}=\frac{32\Omega^2}{12\Omega}=\frac{8}{3}\Omega\approx2.67\Omega$.

Step7: Solve part 2b

First, for two $6\Omega$ resistors, $R_{12}=\frac{6\Omega\times6\Omega}{6\Omega + 6\Omega}=3\Omega$. Then, combining with the $2\Omega$ resistor, $R_{total}=\frac{3\Omega\times2\Omega}{3\Omega + 2\Omega}=\frac{6\Omega^2}{5\Omega}=1.2\Omega$.

Step8: Solve part 2c

First, for $1\Omega$ and $2\Omega$ resistors, $R_{12}=\frac{1\Omega\times2\Omega}{1\Omega + 2\Omega}=\frac{2}{3}\Omega$. Then, combining with the $3\Omega$ resistor, $R_{total}=\frac{\frac{2}{3}\Omega\times3\Omega}{\frac{2}{3}\Omega+3\Omega}=\frac{2\Omega^2}{\frac{2 + 9}{3}\Omega}=\frac{6}{11}\Omega\approx0.55\Omega$.

Answer:

1a. $4\Omega$
1b. $6\Omega$
1c. $\frac{8}{3}\Omega\approx2.67\Omega$
1d. $\frac{12}{5}\Omega = 2.4\Omega$
2a. $\frac{8}{3}\Omega\approx2.67\Omega$
2b. $1.2\Omega$
2c. $\frac{6}{11}\Omega\approx0.55\Omega$