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. what is the total resistance?
3. if the total resistance in figure 5 - 67 is 17.4 kω, what is the value of r5?

figure 5 - 67
+ - r1 r2
5.6 kω 1.0 kω
r3
2.2 kω

• - r4 r5

4.7 kω ?

Explanation:

Step1: Recall series - resistance formula

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

Step2: Solve for twelve 5.6 kΩ resistors in series

For twelve $R = 5.6\ k\Omega$ resistors in series, $R_{total}=12\times R$. Substituting $R = 5.6\ k\Omega$, we get $R_{total}=12\times5.6\ k\Omega=67.2\ k\Omega$.

Step3: Solve for six 56 Ω, eight 100 Ω and two 22 Ω resistors in series

$R_{total}=6\times56\Omega + 8\times100\Omega+2\times22\Omega$. First, calculate each product: $6\times56\Omega = 336\Omega$, $8\times100\Omega = 800\Omega$, $2\times22\Omega = 44\Omega$. Then sum them up: $R_{total}=336\Omega + 800\Omega+44\Omega=1180\Omega$.

Step4: Solve for $R_{5}$ in Figure 5 - 67

Given $R_{total}=17.4\ k\Omega$, and $R_{1}=5.6\ k\Omega$, $R_{2}=1.0\ k\Omega$, $R_{3}=2.2\ k\Omega$, $R_{4}=4.7\ k\Omega$. Using $R_{total}=R_{1}+R_{2}+R_{3}+R_{4}+R_{5}$, we can solve for $R_{5}$ as $R_{5}=R_{total}-(R_{1}+R_{2}+R_{3}+R_{4})$. First, calculate $R_{1}+R_{2}+R_{3}+R_{4}=5.6\ k\Omega+1.0\ k\Omega + 2.2\ k\Omega+4.7\ k\Omega=13.5\ k\Omega$. Then $R_{5}=17.4\ k\Omega - 13.5\ k\Omega = 3.9\ k\Omega$.

Answer:

1. $67.2\ k\Omega$
2. $1180\Omega$
3. $3.9\ k\Omega$