Question

1. calculate (r_t) for each circuit of figure 5 - 66.

Explanation:

Step1: Recall series - parallel resistance formula

For resistors in series, $R_{total}=R_1 + R_2+\cdots+R_n$. For resistors in parallel, $\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}$.

Step2: Analyze circuit (a)

First, the $1.0\ k\Omega$ and $2.2\ k\Omega$ resistors are in series. Their equivalent resistance $R_{s1}=1.0\ k\Omega + 2.2\ k\Omega=3.2\ k\Omega$. Then $R_{s1}$ is in parallel with $5.6\ k\Omega$. So $\frac{1}{R_T}=\frac{1}{3.2\ k\Omega}+\frac{1}{5.6\ k\Omega}=\frac{5.6 + 3.2}{3.2\times5.6\ k\Omega}=\frac{8.8}{17.92\ k\Omega}$, and $R_T=\frac{17.92}{8.8}\ k\Omega\approx2.04\ k\Omega$.

Step3: Analyze circuit (b)

The $4.7\ \Omega$ and $1.0\ \Omega$ resistors are in series, $R_{s2}=4.7\ \Omega+ 1.0\ \Omega = 5.7\ \Omega$. The $10\ \Omega$ and $12\ \Omega$ resistors are in series, $R_{s3}=10\ \Omega + 12\ \Omega=22\ \Omega$. Then $R_{s2}$ and $R_{s3}$ are in parallel. $\frac{1}{R_T}=\frac{1}{5.7\ \Omega}+\frac{1}{22\ \Omega}=\frac{22 + 5.7}{5.7\times22\ \Omega}=\frac{27.7}{125.4\ \Omega}$, and $R_T=\frac{125.4}{27.7}\ \Omega\approx4.53\ \Omega$.

Step4: Analyze circuit (c)

The $1.0\ M\Omega$ and $560\ k\Omega$ resistors are in series. $R_{s4}=1.0\ M\Omega+0.56\ M\Omega = 1.56\ M\Omega$. The $5.6\ M\Omega$ and $680\ k\Omega$ resistors are in series. $R_{s5}=5.6\ M\Omega + 0.68\ M\Omega=6.28\ M\Omega$. Then $R_{s4}$, $R_{s5}$ and $10\ M\Omega$ are in parallel. $\frac{1}{R_T}=\frac{1}{1.56\ M\Omega}+\frac{1}{6.28\ M\Omega}+\frac{1}{10\ M\Omega}=\frac{6.28\times10+1.56\times10 + 1.56\times6.28}{1.56\times6.28\times10\ M\Omega}$. After calculation, $R_T\approx0.89\ M\Omega$.

Answer:

(a) $R_T\approx2.04\ k\Omega$; (b) $R_T\approx4.53\ \Omega$; (c) $R_T\approx0.89\ M\Omega$