Question

1. what is the total resistance of twelve 5.6 kω resistors in series?
2. six 56 ω resistors, eight 100 ω resistors, and two 22 ω resistors are all connected in series.

Explanation:

Step1: Recall series - resistance formula

The formula for the total resistance $R_{total}$ of resistors in series is $R_{total}=\sum_{i = 1}^{n}R_{i}$, where $R_{i}$ are the individual resistances and $n$ is the number of resistors.

Step2: Solve for the first - part

For twelve $5.6k\Omega$ resistors in series, $R_{total1}=12\times5.6k\Omega$.
$R_{total1}=12\times5.6\times10^{3}\Omega = 67.2\times10^{3}\Omega=67.2k\Omega$.

Step3: Solve for the second - part

For six $56\Omega$ resistors, eight $100\Omega$ resistors, and two $22\Omega$ resistors in series:
The sum of the resistances of six $56\Omega$ resistors is $R_1 = 6\times56\Omega=336\Omega$.
The sum of the resistances of eight $100\Omega$ resistors is $R_2 = 8\times100\Omega = 800\Omega$.
The sum of the resistances of two $22\Omega$ resistors is $R_3=2\times22\Omega = 44\Omega$.
Then $R_{total2}=R_1 + R_2+R_3=336\Omega+800\Omega + 44\Omega=1180\Omega$.

Answer:

The total resistance of twelve $5.6k\Omega$ resistors in series is $67.2k\Omega$. The total resistance of six $56\Omega$, eight $100\Omega$, and two $22\Omega$ resistors in series is $1180\Omega$.