a. Applying Eq. (6.1) gives

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdot \cdot \cdot +\frac{1}{R_N}            (6.1)

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

Substituting gives            \frac{1}{4\Omega }=\frac{1}{10\Omega }+\frac{1}{20\Omega }+\frac{1}{R_3}+\cdot \cdot \cdot +\frac{1}{R_N}

so that                                0.25S=0.1S+0.05S+\frac{1}{R_3}

and                                      0.25S=0.15S+\frac{1}{R_3}

with                                     \frac{1}{R_3}=0.1S

and                                      R_3=\frac{1}{0.1S}=10\Omega

b. Using Ohm’s law gives

E=V_1=I_1R_1=(4A)(10\Omega )=40V

c.                                         I_s=\frac{E}{R_T}=\frac{40V}{4\Omega }=10A

d. Applying Ohm’s law gives

I_2=\frac{V_2}{R_2}=\frac{E}{R_2}=\frac{40V}{20\Omega }=2A