Problem 4

4) a)
Between F and H, we have the equivalent resistance given by
$\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{5} + \frac{1}{10}$
$R_{eq} = 2.5 \Omega$

b)
Between F and G, we find the equivalent resistance in three steps. First, let's find an R_intermediate for the $10\Omega$ and $5\omega$ resistors.
$\frac{1}{R_{int}} = \frac{1}{10} + \frac{1}{5}$
$R_{int} = \frac{10}{3} \Omega$
Now, we can combine the resistance we just found and the $5\Omega$ resistor in series with it.
$R_{int2} = \frac{10}{3} + 5$
$R_{int2} = \frac{25}{3} \Omega$
Finally, we can bring it all together.
$\frac{1}{R_{eq}} = \frac{25}{3} + \frac{1}{5}$
$R_{eq} = \frac{25}{8} \Omega$
$R_{eq} = 3.13 \Omega$